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wo 


^  O  bMt&oW  rtfiV  .Cl 

1  sill  o]  bfoa  SK!  o!  :l:->&i  mth  bnj" 

*        r       • 

1;.'>o  fnfr  £t3jffw  amto]  no  dbfitT  ; 


A  SHORT  COURSE 


IN 


GRAPHIC    STATICS 

FOR   STUDENTS   OF 
MECHANICAL   ENGINEERING 


BY 
WILLIAM    LEDYARD    CATHCART 

MEMBER    AMERICAN    SOCIETY   OF    NAVAL   ENGINEERS,    SOCIETY   OF 

NAVAL  ARCHITECTS   AND  MARINE  ENGINEERS,    THE  FRANKLIN 

INSTITUTE,   AMERICAN    SOCIETY   OF   MECHANICAL 

ENGINEERS 

AND 

J.    IRVIN    CHAFFEE,   A.M. 

PROFESSOR   OF   MATHEMATICS,    WEBB'S   ACADEMY   OF  NAVAL 

ARCHITECTURE   AND    MARINE    ENGINEERING,    MEMBER 

SOCIETY   OF   NAVAL  ARCHITECTS   AND   MARINE 

ENGINEERS 


58   ILLUSTRATIONS 


NEW   YORK 
D.   VAN    NOSTRAND    COMPANY 

23  MURRAY  AND  27  WARREN  STREETS 
IQII 


COPYRIGHT,   1911,  BY 
D.  VAN  NOSTRAND  COMPANY 


NotiBOOtJ 

J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE 

THE  purpose  of  this  book  is  to  provide  students  of 
mechanical  engineering  with  a  brief  course  in  Graphic 
Statics  which  will  serve  when  the  time  to  be  devoted 
to  this  subject  is  short.  Owing  to  the  necessary  limita- 
tions as  to  size,  the  treatment  has  been  restricted  mainly 
to  the  properties  and  general  uses  of  the  force  and  equi- 
librium polygons,  these  polygons  being  sufficient  for 
the  solution  of  most  of  the  problems  met  in  practice 
by  mechanical  engineers.  While  the  design  of  trusses 
is,  in  general,  the  duty  of  the  civil  engineer,  some  atten- 
tion has  been  given  this  subject,  since  such  constructions 
as  the  Warren  girder  for  an  overhead  crane,  the  walking 
beam  of  an  engine,  etc.,  fall  under  this  classification. 
Examples,  in  full  detail,  have  been  included,  so  far  as 
space  would  admit,  since  one  good  example  is  often  of 
more  service  in  instruction  than  many  pages  of  theoreti- 
cal investigation. 

The  discussion  of  principles,  as  given  in  this  book, 
is  largely  a  summary  of  similar  portions  of  the  authors' 
creatise,  "The  Elements  of  Graphic  Statics,"  although 
there  have  been  some  minor  additions.  With  the  latter 
and  the  examples,  the  new  material  comprises  nearly 
two-thirds  of  the  text. 

NEW  YORK,  October  i,  1911. 


255809 


CONTENTS 

CHAPTER   I 

PAGE 

FORCE  AND  EQUILIBRIUM  POLYGONS i 

Article  i.  Force  Triangle  —  2.  Force  Polygon — 3.  Equi- 
librium Polygon  —  4.  Conditions  of  Equilibrium. 

Example  i.  Bellcrank  —  2.  Pawl  and  Ratchet  — 
3.  Ratchet-rack  —  4.  Stationary  Engine  —  5.  Pillar  Crane 

—  6.  Sheer  Legs. 

CHAPTER   II 

TRUSSES:   STRESS  DIAGRAMS 23 

Article  5.  Framed  Structures  —  6.  Stress  Diagrams. 
Example  7.  Roof  Truss,  Dead  Load  — 8.  Roof  Truss, 
Wind  Load  —  9.  Crane  Truss. 

CHAPTER   III 

STATIONARY  LOADS:   SHEARS  AND  MOMENTS      .        .        .41 

Article  7.  Beams  —  8.  Vertical  Shear  —  9.  Bending  Mo- 
ment—  10.  Resisting  Moment — n.  Shear  and  Moment 
Diagrams — 12.  Moment  Scale — 13.  Twisting  Moments 
— 14.  Bending  and  Twisting  Moments  combined :  Equiva- 
lent Bending  and  Twisting  Moments. 

Example  10.  I-Beam,  Uniformly  Distributed  Load  — 
ii.  Locomotive  Side  Rod  with  Uniform  Load  due  to  Cen- 
trifugal Force —  12.  Girder  Stay  with  Stresses  produced 
by  its  supporting  a  Continuous  Beam  under  Uniform  Load 

—  13.  Counter  Shaft:    Twisting  and  Bending  Combined 
— 14.  Centre  Crank  Shaft :  Twisting  and  Bending  Com- 
bined. 


VI  CONTENTS 


CHAPTER   IV 

PAGE 

LIVE  LOADS:   SHEARS  AND  MOMENTS 73 

Article  15.  Variation  of  Live  Load  Shear  at  Sections  to 
the  Left  of  the  Load —  16.  Influence  Diagrams  :  Influence 
Lines  —  17.  Variation  of  Live  Load  Shear  at  Any  Given 
Section  of  a  Beam — 18.  Maximum  Live  Load  Shear  — 
19.  Counterbracing  —  20.  Variation  of  Bending  Moment 
at  Any  Given  Section  of  a  Beam  —  21 .  Maximum  Bending 
Moments  due  to  Live  Loads  —  22.  Live  Load  Stresses  in 
Trusses  and  Plate  Girders. 

Example  15.  Plate  Girder  Bridge  with  Locomptive 
Wheel  Loads  :  Maximum  Moments  and  Shears  — 16.  War- 
ren Girder  for  Overhead  Crane:  Maximum  Stresses  — 
17.  Pratt  Truss:  Uniform  Live  Load. 


CHAPTER  V 

CENTRE  OF  GRAVITY:   MOMENT  OF  INERTIA        .        .        .no 

Article  23.  Centre  of  Gravity  —  24.  Centroid  of  Two 
Parallel  Forces  —  25 .  Centroid  of  Complanar  Parallel 
Forces  whose  Points  of  Application  are  Complanar  with 
All  of  the  Forces,  but  are  not  in  a  Straight  Line  — 
26.  Moment  of  Inertia:  Radius  of  Gyration  —  27.  Mo- 
ment of  Inertia  of  a  System  of  Complanar  Parallel  Forces 
—  28.  Parallel  Axes  of  Inertia,  One  passing  through  the 
Centroid  —  29.  Moment  of  Inertia  of  an  Area. 

Example  18.  Centroid  of  Parallel  Forces  due  to  Loco- 
motive Wheel  Loads — 19.  Centre  of  Gravity  of  Bulb 
Angle  —  20.  Centre  of  Gravity  of  a  Partial  Area  —  21 .  Mo- 
ment of  Inertia  of  the  Cross-section  of  a  Deck  Beam :  Ap- 
proximate Method  —  22.  Accurate  Determination  of  the 
Moment  of  Inertia  of  an  Area  about  an  Axis  passing 
through  its  Centre  of  Gravity. 


CONTENTS  Vll 


CHAPTER   VI 

PACK 

FRICTION 134 

Article  30.  Friction  —  31.  Friction  of  Plane  Surfaces: 
Friction  Cone  —  32.  Friction  of  Screw  Threads — 33.  Pivot 
and  Collar  Friction  —  34.  Journal  Friction  :  Friction  Circle 
—  35.  Link  Connections:  Friction  Axis  —  36.  Chain 
Friction:  Resistance  of  Ropes  to  Bending  —  37.  Belt 
Gearing  —  38.  Friction  of  Gear  Teeth. 

Example  23.  Friction  of  Stationary  Engine  —  24.  Fric- 
tion of  Screw  Jack  —  25.  Pulley  Blocks  :  Relation  of  Load 
and  Power  —  26.  Spur  Gears :  Relation  of  Load  and 
Power. 


GRAPHIC    STATICS 


CHAPTER    I 

FORCE    AND    EQUILIBRIUM    POLYGONS 

The  Force  Triangle,  its  extension,  the  Force  Poly- 
gon, and  the  Equilibrium  or  Funicular  Polygon  form 
the  foundations  of  all  important  applications  in  practice 
of  the  science  of  Graphic  Statics.  Consider  first : 

1.  Force  Triangle.  In  Fig.  i,  the  weight  W '=  300 
pounds  and  the 
weights  Wl  and 
W<i  are  con- 
nected at  O  by 
cords,  two  of 
which  pass  over 
frictionless  sup- 
ports at  b  and  c. 
The  system  is 
in  equilibrium 
and  W  is  at  rest, 
being  held  from 
vertical  or  hori- 
zontal movement  by  the  tensile  stresses  7i  and  T2  in 
the  supporting  cords,  which  stresses  are  equal  to  their 


FIG. 


2  .GRAPHIC    STATICS 

corresponding  weights  W±  and  W^  respectively.  It 
is  required  to  find  the  magnitudes  of  these  weights. 

On  any  scale  —  say,  200  pounds  to  the  inch  —  lay  off 
the  line  de  —  T  =  W  =  300  pounds,  parallel  to  the  line 
of  action  Oa  of  the  stress  or  force  T.  Draw  df  and 
eft  parallel  respectively  to  Ob  and  Oc.  Then,  meas- 
ured by  the  same  scale,  fd  —  T^  —  W^  —  224  pounds 
and  ef  —  T2  =  W2  =  184  pounds. 

The  triangle  def  is  the  force  triangle  for  the  forces 
T,  Tv  and  T2.  The  direction  of  these  forces  is  indi- 
cated by  the  arrowheads.  It  will  be  seen  that,  in  pass- 
ing around  the  force  triangle,  the  three  forces  have  the 
same  direction,  which  is  the  case  always  when  the  sys- 
tem is  in  equilibrium.  On  the  other  hand,  the  resultant 
of  7\  and  T2  is  a  force  T'  =  T,  applied  at  O,  and  this 
would  be  shown  by  the  line  ed,  whose  direction,  as 
given  by  the  upper  arrowhead,  is  the  reverse  of  those 
of  7*!  and  T2,  in  passing  around  the  triangle.  Similarly, 
a  force  directed  from  d  to  f  would  be  the  resultant  of 
T  and  Tz,  and  one  from  f  to  e,  the  resultant  of  T  and 
7\.  These  principles  are  general,  i.e.,  if,  in  any  force 
triangle,  the  forces  have  the  same  direction  in  passing 
around  the  triangle,  the  system  is  in  equilibrium  ;  and, 
second,  the  force  represented  by  any  side  of  a  force  tri- 
angle is  the  resultant  of  the  forces  represented  by  the 
other  two  sides,  if  its  direction,  in  passing  around  the 
triangle,  is  the  reverse  of  theirs.  The  system,  which 
consists  of  two  forces  and  their  resultant,  is  not  then 
in  equilibrium,  since  a  resultant,  unbalanced  force 
exists. 


FORCE   AND   EQUILIBRIUM   POLYGONS 


2.  Force  Polygon.  The  principles  governing  the  con- 
struction of  the  force  polygon  are  an  extension  of  those 
of  the  force  triangle.  Thus,  Fig.  2,  let  there  be  four 
forces  7\  •  •  •  T4  acting  at  the  angles  shown,  the  mag- 
nitude of  each  being 
100  pounds,  and  all 
meeting  at  the  point 
O  to  support  the 
weight  W,  the  mag- 
nitude of  which  is 
required. 

On  a  scale  of  100 
pounds  to  the  inch, 
lay  off  ab,  be,  cd,  and 
det  parallel  respec- 
tively to  the  lines  of 
action  of  Tlt  T2,  T3, 
and  r4.  Then,  ae  = 
T'=  270  pounds  is 
the  resultant  of  the 
four  forces,  and  is 
therefore  equal  in  magnitude  to  W  and  opposite  in 
direction  to  Tt  the  stress  produced  by  W.  This  follows 
since,  if  the  polygon  be  divided  into  triangles  by  the 
dotted  diagonals  shown,  ac  is  the  resultant  of  ab 
and  be,  ad  is  the  resultant  of  ac  and  cd,  and  ae=  T' 
is  the  resultant  of  ad  and  de  and  therefore  of  the  four 
forces.  The  principles  established  for  the  force  triangle 
apply  also  to  the  force  polygon.  With  either  the  tri- 
angle or  the  polygon,  the  sides  representing  forces  may 


FIG.  2 


4  GRAPHIC    STATICS 

be  drawn  in  any  order,  the  result  being  always  the  same, 
since  the  force  polygon  is  essentially  but  the  graphic  addi- 
tion of  forces,  i.e.,  the  horizontal  and  vertical  components 
of  the  resultant  of  a  system  of  forces  are,  respectively, 
the  algebraic  sum  of  the  horizontal  and  vertical  compo- 
nents of  the  forces  forming  the  system. 

The  forces  considered  in  Figs.  I  and  2  are  concurrent, 
i.e.,  their  lines  of  action  meet  at  a  common  point  O. 
With  non-concurrent  forces,  the  lines  of  action  may  be 
either  inclined  to  each  other  or  parallel,  but  in  neither 
case  have  they  a  common  point  of  intersection.  The 
force  polygon  gives  only  the  magnitude  of  a  force  and 
the  inclination  of  its  line  of  action ;  it  does  not  fix  the 
location  of  that  line  within  the  body  or  area  considered. 
For  these  reasons,  such  polygons  can  determine  the 
equilibrium  of  concurrent  forces  only,  the  location  of 
whose  lines  of  action  is  governed  by  the  fact  that  they 
must  all  intersect  at  one  point.  With  non-concurrent 
forces,  the  same  force  polygon  may  be  drawn  for  forces 
of  the  same  magnitude  and  inclination,  when  these 
forces  are,  and  are  not,  in  equilibrium.  Thus,  Fig.  I,  if 
the  force  T2  be  transferred  to  the  new  line  of  action 
O'c1 ,  the  system  becomes  non-concurrent,  is  no  longer 
in  equilibrium,  and  yet  the  force  triangle  edf  remains  the 
same. 

3.  Equilibrium  Polygon.  In  order  to  determine  the 
equilibrium  of  a  system  of  non-concurrent  forces,  both 
the  force  and  equilibrium  (funicular)  polygons  must  be 
drawn.  Thus,  let  Fig.  3  represent  non-concurrent  forces 


FORCE    AND    EQUILIBRIUM   POLYGONS 


Pl"-PB  applied  to  a  rigid  body,  the  magnitude,  direc- 

tion, and  line  of  action  of  each  force  being  known  and 

the   system   being  in  equilibrium.     The  force  polygon 

for  these  forces  is  a  •••  e,  as  shown  below.     Starting  at 

any  point  on  any  line 

of    action,    draw   the 

equilibrium     polygon 

A  ">  E,  of  any  shape, 

the     only    conditions 

being   that   its    sides 

Si'-  S&  shall  be  the 

same    in    number   as 

the   forces,   and   that 

each     pair    of    sides 

shall  intersect  on  the 

line  of  action  between 

them.     In    the    force 

polygon,     draw     the 

rays   S^-S^   parallel 

to    their    correspond- 

ing sides  in  the  equilibrium  polygon.     It  will  be  found 

that  these  rays  all  intersect  at  a  common  point  (7,  called 

the  pole. 

Now,  consider  the  equilibrium  polygon  as  a  jointed 
frame,  composed  of  rigid  bars  pivotally  connected  at 
the  joints  A  --  E,  this  frame  being  in  equilibrium  under 
the  action  of  the  external  forces  P1-"P&  and  the  stresses 
S1  ~-  S5  in  the  sides.  Since  the  frame  is  in  equilibrium, 
each  joint  must  also  be  in  equilibrium.  Therefore,  for 
joint  A,  the  triangle  Oae  is  the  force  triangle  for  the 


FlG  3 


6  GRAPHIC   STATICS 

force  Pl  and  the  stresses  Sl  and  S5.  The  direction  of 
P1  is  given,  and  this  determines  the  direction  of  Sl  and 
S6,  with  regard  to  joint  A,  as  shown  by  the  arrowheads 
in  the  force  polygon,  the  three  forces  being  in  equilib- 
rium and  hence  having  the  same  direction  in  passing 
around  the  triangle.  Transferring  these  directions  to 
the  equilibrium  polygon,  it  will  be  seen  that,  since  S± 
and  S5  both  act  away  from  joint  A,  they  must  both  be 
tensile  stresses. 

The  arrowheads  just  drawn  on  the  rays  relate  only  to 
the  direction  of  the  stresses  with  regard  to  joint  A.  Each 
joint  must  be  treated  separately  in  this  respect.  Thus, 
at  joint  B  the  direction  of  P%  is  known  and  also  that  of 
Sv  since  the  latter  has  already  been  found  to  be  tensile 
and  therefore  pulls  away  from  joint  B,  reversing  its 
direction  in  the  force  polygon  as  given  for  joint  A.  As 
the  stress  S2  must  have  the  same  direction  in  the  tri- 
angle Obat  it  follows  that  it  is  tensile.  SB  and  54.are 
also  found  to  be  tensile  stresses.  If  any  stress  had 
acted  toward  any  joint,  it  would  have  been  compressive. 
The  character  of  a  stress,  as  to  tension  or  compression, 
is  thus  given  by  its  direction  in  the  force  polygon,  when 
that  direction  is  transferred  to  the  line  of  action  of  the 
stress  at  the  joint  in  question ;  its  magnitude  is  shown 
by  its  length  in  the  force  polygon,  as  measured  by  the 
force-scale  of  the  latter. 

As  stated,  Fig.  3  represents  a  system  of  forces  in 
equilibrium.  That  the  equilibrium  polygon  must  close 
for  equilibrium  to  exist,  may  be  shown  by  transferring 
the  force  PR  to  the  new  line  of  action  E1  E" .  The 


FORCE  AND  EQUILIBRIUM  POLYGONS         7 

system  will  then  be  no  longer  in  equilibrium,  the  side 
DE  will  end  at  E1 ,  the  side  AE  will  be  prolonged  to  E", 
and  the  polygon  will  not  close.  In  any  force  polygon, 
the  rays  all  meet  at  the  pole  O,  because  each  ray  is 
common  to  two  adjacent  force  triangles.  It  is  evident, 
also,  that  an  infinite  number  of  equilibrium  polygons  can 
be  drawn  for  the  same  system  of  external  forces,  since 
two  consecutive  rays  simply  show  the  magnitude  and 
inclination  of  the  two  stresses  which,  at  these  angles  of 
inclination,  will  hold  the  external  force  at  the  corre- 
sponding joint  in  equilibrium.  If  the  inclination  of 
the  sides  of  the  equilibrium  polygon  be  changed,  the 
magnitude  of  these  stresses  will  be  altered  correspond- 
ingly. Hence,  the  pole  may  fall  at  any  point  within  or 
without  the  force  polygon  ;  or,  reversing  the  operation, 
any  point  may  first  be  selected  as  a  pole,  the  rays  drawn 
to  the  vertices  of  the  force  polygon,  and  then  the  equi- 
librium polygon  constructed  with  its  sides  parallel  to 
these  rays.  Finally,  the  equilibrium  polygon  may  be 
substituted  for  any  rigid  body  on  which  a  system 
of  forces  —  as  Pl  .  .  .  P5  —  acts,  since  a  force  may  be 
considered  as  applied  at  any  point  in  its  line  of 
action. 

When  the  lines  of  action  of  the  forces  are  parallel, 
the  same  principles  apply,  except  that  the  force  polygon 
now  becomes  a  straight  line.  Thus,  let  Fig.  4  represent 
a  simple  beam  of  24  feet  span,  supported  at  A  and  B, 
loaded  with  weights  Pl  and  Pz  of  300  and  200,  pounds, 
respectively,  and  having  a  vertical,  upward  thrust, 
/>2=  I oo  pounds.  It  is  required  to  determine  the  magni- 


8 


GRAPHIC    STATICS 


tudes  of  the  vertical  reactions  Rl  and  R2  at  the  left  and 
right  supports,  respectively. 

On  the  load-line  ad  lay  off,  to  any  convenient  scale, 
the  downward* force  Pl  =  300  pounds  =  ab,  the  upward 
force  P2  =  100  pounds  =  be,  and  -the  downward  force 

P3  =  200    pounds  =  cd. 

pe  5'  -4-  4'  -H«—  --»)*  4-»j       Xake  any  point  (9  as  the 

-fi  pole,  and  draw  the  rays 
<9#,  <9£,  6><:,  and  Od. 
Starting  at  any  point  F 
on  the  line  of  action  of 
Plt  draw  the  equilibrium 
polygon  whose  sides  FG, 
FH,  HK,  and  KL  are 
parallel,  respectively,  to 
the  rays  Oa,  Ob,  Oc,  and 
Od.  Finally,  draw  the 
closing  side  LG,  and  the 
corresponding  ray  Oe. 
Then,  measured  to  the 
load-scale,  de  —  R^—\^\.J 
pounds,  ea  =  R1  =  208.3  pounds,  and  the  closed  force 
polygon  is  the  straight  line  abcdea. 

It  will  be  seen  that,  in  selecting  a  pole  and  drawing 
the  rays  and  the  corresponding  equilibrium  polygon, 
we  have  simply  assumed  two  stresses  at  each  joint  of 
the  latter  polygon  which  will  hold  the  external  force 
at  that  joint  in  equilibrium,  the  lines  of  action  of  the 
stresses  being  determined  by  the  inclination  of  the 
rays.  Thus,  at  joint  F,  the  assumed  stresses  have 


FIG.  4 


FORCE  AND  EQUILIBRIUM  POLYGONS         9 

the  lines  of  action  FG  and  FH  and  their  magnitudes 
are  equal  to  Oa  and  Ob,  respectively,  when  measured 
on  the  scale  of  the  force  polygon.  Since  the  three 
forces  Pv  Oa,  and  Ob  have  the  same  direction  in 
passing  around  the  triangle  Oab,  the  system  at  joint 
F  is  in  equilibrium.  The  same  principles  apply  to 
each  of  the  other  joints.  The  direction  of  the  closing 
line  LG  of  the  polygon  determines  that  of  the  ray  Oe, 
and  hence  the  magnitudes  of  the  two  reactions,  since 
the  latter  are,  in  effect,  vertical,  upward  forces,  held 
in  equilibrium  by  the  stress  in  LG  or  Oe  acting  with 
that  in  FG  or  Oa  for  Rlt  and  with  that  in  KL  or  Od 
for  R^ 

4.  Conditions  of  Equilibrium. — The  forces  considered 
in  graphical  processes  are  complanar,  i.e.,  their  lines  of 
action  all  lie  in  the  same  plane.  As  to  such  forces,  it 
will  be  seen  from  the  foregoing  that,  for  the  equili- 
brium of: 

(a)  Concurrent  forces,  the  force  polygon  must  close ; 
and,  conversely,  if  the  force  polygon  closes,  the  system 
is  in  equilibrium. 

(b)  Non-concurrent  forces,  both  the  force  and  equilib- 
rium polygons  must  close ;  conversely,  if  the  force  poly- 
gon and  any  equilibrium  polygon  close,  the  system  is  in 
equilibrium. 

In  the  graphical  analysis  of  the  forces  acting  on  a 
moving  body,  the  latter  is  assumed  to  be  in  momentary 
equilibrium  under  the  action  of  the  driving  forces,  the 
reactions  at  journals  or  other  supports,  and  the  forces 


10  GRAPHIC    STATICS 

due  to  resistances  to  motion.  The  forces  or  stresses 
thus  determined  apply  only  to  the  given  position  of  the 
moving  parts. 

When  it  is  desired  to  use  graphical  methods  with 
forces  whose  lines  of  action  lie  in  different  planes,  the 
components  of  these  forces  in  one  common  plane  for  all 
should  be  found.  These  components  can  then  be  treated 
as  a  complanar  system. 

EXAMPLES 

In  the  following  examples  of  the  application  of  the 
force  polygon,  the  friction  of  journals  and  other  bear- 
ings is  not  considered.  When  a  body  rests  on  another, 
as  a  crosshead  on  its  bearing,  there  is  a  reacting  pressure 
from  the  bearing,  equal  and  opposite  to  the  downward 
pressure  upon  it.  While  the  parts  are  at  rest,  this  re- 
action is  perpendicular  to  the  surfaces  in  contact ;  when 
they  move,  the  reaction  would  still  be  normal  to  these 
surfaces,  if  there  were  no  friction  (Art.  30) ;  by  friction, 
its  line  of  action  is  diverted  from  the  normal  by  the 
amount  of  the  angle  of  friction,  as  will  be  explained  later. 
The  total  reaction,  which  is  the  resultant  of  the  indefinite 
number  of  indefinitely  small  reactions  at  all  points  of 
the  bearing,  is  assumed  theoretically  to  act  at  the  centre 
of  the  bearing  surface ;  in  practice,  this  is  only  approxi- 
mately true,  since  uneven  wear  makes  the  bearing  pres- 
sure variable  throughout. 

i.  BELL-CRANK.  Taking  the  simplest  example  first, 
consider  the  bell-crank  AOB,  Fig.  5,  journalled  at  O, 
driven  by  the  link  CA  with  a  force  /\  =  100  pounds, 


FORCE    AND   EQUILIBRIUM   POLYGONS 


II 


and  driving  the  link  BD,  whose  resistance  P2  is  to  be 
determined.  The  bell-crank  is  assumed  to  be  at  mid- 
stroke. 

The  crank  is  in  momentary  equilibrium  under  the 
action  of  the  forces  Plf  P%  and  the  reacting  pressure  R 
from  the  journal,  which 
reaction  is  opposed  to  the 
resultant  of  the  two  forces. 
Since  equilibrium  exists 
and  the  forces  are  concur- 
rent, the  three  lines  of 
action  must  meet  at  a  com- 
mon point  E,  which  is  one 
point  in  the  line  of  action 
of  the  reaction  R.  The 
other  point  necessary  is 
determined  by  the  fact 
that,  as  friction  is  disre- 
garded, the  reaction  is  normal  to  the  contact-surfaces  of 
the  bearing,  and  hence  its  line  of  action  passes  through 
the  centre  O.  Therefore,  in  the  force  triangle  abc,  draw 
ab—P^—  loo  pounds  and  lay  off  be  and  ca  parallel, 
respectively,  to  EO  and  BD.  Then,  on  the  same  scale, 
be  =  R  =  1 80  pounds  and  ca  —  P^—  1 50  pounds. 

2.  PAWL  AND  RATCHET.  Figure  6  represents  a  link 
AB,  driving  with  a  force  Pl  =  50  pounds  the  crank  BO 
loose  on  the  shaft  O  and  carrying  the  pawl  C,  which 
engages  the  ratchet  D  keyed  to  the  shaft.  It  is  re- 
quired to  determine  the  resistance  of,  and  the  reaction 
upon,  the  pawl-crank. 


FIG.  5 


12 


GRAPHIC   STATICS 


The  crank  is  in  momentary  equilibrium  under  the 
action  of  the  driving  force  Plt  the  backward  thrust  P%, 
due  to  the  resistance  of  the  ratchet,  and  the  reaction  R 
from  the  shaft,  which  reaction  is  equal  and  opposite  to 
the  resultant  of  Pl  and  P2. 

Since  equilibrium  exists  and 
the  forces  are  concurrent,  the 
lines  of  action  of  the  latter  meet 


FIG.  6 


FIG.  7 


at  the  common  point  B  To  draw  the  force  triangle, 
lay  off  ab  =  Pl  =  50  pounds,  parallel  to  AB  and  set  off 
be  and  ca,  parallel  respectively  to  EB  and  BO.  Then, 
be  —  P2  =  70  pounds  and  ca  =  R  =  50  pounds. 

3.  RATCHET-RACK.  Figure  7  shows,  in  partial  eleva- 
tion, the  double  ratchet-rack,  a  mechanism  sometimes 
used  as  a  jack  for  raising  light  weights.  The  rack  A 
has  ratchet  teeth  on  two  opposite  sides,  which  are  alter- 


FORCE   AND   EQUILIBRIUM   POLYGONS  13 

nately  engaged  by  the  two  oppositely  inclined  pawls, 
B  and  Ct  mounted  on  the  lever  £>,  which  is  pivoted  at 
E  on  the  framing  (not  shown).  The  rack  is  guided  by 
two  central  ribs  which  reciprocate  in  guide-grooves 
formed  in  the  frame.  In  Fig.  7,  the  driving  lever  is 
assumed  to  be  at  mid-stroke,  with  the  pawl  B  lifting  the 
rack  and  the  weight  W  carried  by  the  latter. 

The  thrust  of  the  driving  pawl  is  applied  so  near  the 
centre  line  of  the  rack  that  its  leverage  and  the  conse- 
quent tendency  to  cant  the  rack  are  but  slight.  Hence, 
the  reactions  R1  and  R^  of  the  bearings  press  toward 
the  left  and  from  the  right  sides  of  the  grooves.  These 
reactions  are  assumed  to  act  from  the  respective  centres, 
F  and  G,  of  the  bearing  surfaces ;  the  reaction  shown 
at  each  bearing  is  taken  as  that  on  both  ribs. 

The  rack  is  in  momentary  equilibrium  under  the 
action  of  the  weight  W,  the  two  reactions,  and  the  thrust 
P  of  the  driving  pawl.  There  are  thus  four  forces 
which  are  not  concurrent,  since  the  two  reactions  are 
parallel.  The  analysis  of  these  forces  can  be  made 
either  by  the  use  of  the  force  and  equilibrium  polygons 
or  by  the  method  of  resultants,  as  given  below. 

Equilibrium  of  this  system  of  four  forces  can  exist 
only  when  the  resultant  of  any  pair  of  them  is  equal  to 
that  of  the  other  pair,  and  the  two  resultants  are  oppo- 
site in  direction  and  have  the  same  line  of  action.  The 
lines  of  action  of  the  weight  W  and  the  resultant  Rl 
intersect  at  //,  which  is  one  point  on  the  line  of  action 
of  their  resultant ;  similarly,  the  lines  of  action  of  P  and 
R2  meet  at  K,  which  is  a  point  on  the  line  of  action  of 


14  GRAPHIC    STATICS 

their  resultant.  Since  the  two  resultants  must  have  the 
same  line  of  action,  it  is  obvious  that  this  line  is  HK. 

Take  W  as  300  pounds,  and,  in  the  force  polygon, 
lay  off  ab  =  W=  300 ;  draw  be  parallel  to  HK  and  ac 
parallel  to  Rv  Similarly,  from  b  and  c  lay  off  bd 
and  cd,  parallel  respectively  to  KL  and  GK.  Then, 
ac  =  Rl=ioo  pounds,  cd=R^  =  ^Q  pounds,  and  bd 
=  P  =  334  pounds.  While  this  method  would  be  similar 
for  any  distance  between  the  bearings,  and  between  the 
rack  and  the  pin  of  the  driving  pawl,  the  numerical  re- 
sults, as  above,  apply  only  to  the  spacing  shown  in  the 
figure  and  to  the  given  position  of  the  parts. 

4.  STATIONARY  ENGINE.  Let  Fig.  8  represent  dia- 
grammatical ly  the  ordinary  stationary  steam  engine,  the 


FIG.  8 


piston  pressure  P  being  applied  at  the  crosshead  pin  A 
and  transferred  through  the  connecting  rod  AB  to  the 
crank  BC  keyed  to  the  shaft  Cy  which  rotates  in  an  anti- 
clockwise direction. 


FORCE   AND   EQUILIBRIUM  POLYGONS  15 

First,  to  determine  the  reaction  at  the  crosshead  bear- 
ing: disregarding  friction,  this  reaction  will  be  perpen- 
dicular to  the  contact-surfaces  and  therefore  vertical. 
The  crosshead  pin  is  thus  in  equilibrium  under  the  ac- 
tion of  three  forces :  the  piston  pressure  P,  acting  hori- 
zontally and  from  right  to  left ;  the  diagonal,  downward 
resistance  T  of  the  connecting  rod ;  and  the  vertical 
and  upward  reaction  R1  of  the  crosshead  bearing.  Since 
equilibrium  exists,  the  lines  of  action  of  these  three 
forces  intersect  at  a  common  point,  and  this  point  must 
be  A,  as  P  and  T  meet  there.  In  the  force  polygon 
abed  lay  off,  to  any  convenient  scale,  ab  =  P\  draw  ac 
and  be,  parallel  respectively  to  the  lines  of  action  of  T 
and  R^  Then,  on  the  same  scale,  ca—T  and  be  —  Rlt 

Second,  to  find  the  reaction  at  the  shaft  bearing:  to 
ascertain  this,  the  character  of  the  resistance  must  be  as- 
sumed. Suppose  that  a  spur  gear  CD  is  keyed  to  the 
shaft,  and  that  this  gear  meshes  at  D  with  a  similar 
gear  which  it  drives.  There  will  be  evidently  then  at 
D  a  resistance  Q,  acting  from  left  to  right  and  perpen- 
dicular to  the  radius  CD.  The  conditions  are  thus  simi- 
lar to  those  of  Example  I,  since,  as  the  gears  and  crank 
are  keyed  to  the  shaft,  the  three  members  virtually 
form  a  bell-crank  which  is  pressed  to  the  left  at  B  by 
the  force  T,  now  a  driving  force ;  is  forced  to  the  right 
by  the  resistance  Q,  acting  at  D\  and  is  held  from  ris- 
ing out  of  the  bearing  by  the  reaction  R^  of  the  cap  of 
the  latter.  The  lines  of  action  of  Q  and  T  meet  at  5. 
Hence,  for  equilibrium,  that  of  R^  must  be  SC.  In  the 
force  polygon,  we  have  already  found  the  magnitude  of 


i6 


GRAPHIC   STATICS 


the  force  T,  which  now  acts  from  a  to  c.  Therefore, 
from  a  lay  off  ad,  and  from  c  lay  off  cdt  parallel  respec- 
tively to  the  lines  of  action  of  /?2  and  Q.  Then,  cd  =  Q 
and  da  =  Rv 

5.    PILLAR  CRANE.     In  the  crane  shown  diagrammati- 
cally  in  Fig.  9,  AB  is  the  tie,  AC  the  boom,  and  BC  the 


X-T5' 


post  or  pillar.  The 
weight  W  is  sus- 
pended by  a  crane 
hook  from  the  block 
D.  One  end  of  the 
hoisting  rope  is  fast- 
ened to  the  tie-rod 
and  the  other  passes 
around  the  sheaves 
in  blocks  D  and  A 
and  thence  to  the 
hoist  at  E,  the  latter 
being  secured  to  the 
boom.  Let  the  blocks 
be  single-sheaved 
with  a  velocity-ratio 
of  2  and  a  consequent 
tension  T=  W/2  in 
the  hoisting  rope. 
Take  W  as  5  tons 
(10,000  pounds)  and  neglect  the  weight  of  the  crane. 
It  is  required  to  determine  the  stresses  in  the  tie,  boom, 
and  post,  and  to  show  the  general  effect  on  the  two  for- 
mer stresses  of  the  tension  T. 


FORCE    AND    EQUILIBRIUM    POLYGONS  17 

First,  neglect  Tt  i.e.,  assume  W  to  be  simply  hung 
from,  and  supported  by,  the  joint  at  A.  This  joint  is 
then  in  equilibrium  under  the  action  of  Wand  the  loads 
P1  and  P2  on  the  tie  and  boom,  respectively.  In  the 
force  polygon  abc,  lay  off  ab  =  W=  5  tons,  and  draw 
ac  and  be,  parallel  respectively  to  AB  and  AC.  Then, 
ca  —  P1  =  8.8  tons  is  the  tensile  load  on  the  tie,  and  be 
=  P2  =  11.4  tons  is  the  compressive  load  on  the  boom, 
due  to  the  resultant  along  cb  of  Wand  Pv 

Joint  B  is  in  equilibrium  under  the  action  of  the  ten- 
sile force  P!  and  the  stresses  which  its  horizontal  and 
vertical  components  produce.  The  horizontal  compo- 
nent, P3  =  ad=8.3  tons,  acts  to  bend  the  pillar,  and 
this  bending  is  resisted  by  stresses  whose  resultant  S%  is 
equal  and  opposite  to  P%  and  acts  at  B.  The  vertical 
component,  P±  =  dc  —  2.9  tons,  tends  to  stretch  either 
the  post  or  vertical  ties  between  B  and  C.  This  force 
is  resisted  by  an  equal  and  opposite  tensile  stress  54. 

At  joint  C  there  are  acting  the  diagonal  thrust  P2  on 
the  boom,  the  tensile  stress  54,  an  upward  reaction 
R  =  W  on  the  collar-  or  pivot-bearing  of  the  post,  and  a 
horizontal  reaction  R%  from  the  bearing  at  the  base. 
The  horizontal  and  vertical  components  of  P2  are  P6 
and  Pp  respectively.  The  force  polygon  is  cbedc,  in 
which  cb  =  Pz,  t>e=P5  =  P#  ed=R=W,dc=  54,  and 
ce  =  PQ  =  54  -h  R  =  7.9  tons. 

Now,  consider  the  tension  T  in  the  hoisting  rope  be- 
tween the  joint  A  and  the  hoist  E.  This  tension  has 
four  direct  and  indirect  effects :  it  acts  as  a  tensile  force, 
diagonal  and  to  the  left,  on  the  joint  A  ;  as  a  result,  its 


1 8  GRAPHIC    STATICS 

component  parallel  to  the  tie  takes  a  part  of  the  original 
load  on  the  latter,  and  its  component  parallel  to  the 
boom  acts  both  as  an  additional  compressive  load  be- 
tween A  and  the  point  of  attachment  Foi  the  hoist,  and 
as  a  tensile  load  between  F  and  the  lower  end  of  the 
boom  at  C. 

To  lay  out  the  new  force  polygon  for  joint  A,  draw 
af=T=  W/2  parallel  to  EA,  and  from/draw/^7  paral- 
lel to  AB  and  meeting  be  prolonged  at  c'.  The  poly- 
gon is  then  abc'fa  and  c'f=  7.5  tons  is  the  tensile  load 
on  the  tie,  while  be'  =  12.7  tons  is  the  compressive  load 
on  the  boom  between  the  points  F  and  A.  It  will  be 
seen  that  these  changes  in  the  two  loads  amount  simply 
to  adding  the  component  gf  of  T  parallel  to  the  boom  to 
the  compressive  load  on  the  latter,  and  to  deducting  the 
similar  component  ga  from  the  original  tensile  load  on 
the  tie.  Now,  from  c  lay  off  cc"  —fg.  Then,  c"b  =  10 . 1 
tons  is  the  new  compressive  load  on  the  boom  between 
Fand  C.  These  changes  in  the  load  on  the  tie  and  in 
the  thrust  on  the  lower  end  of  the  boom  alter  corre- 
spondingly the  values  of  P3,  P^  P6,  and  P6,  with  their 
resulting  stresses  and  reactions,  although  R  is  still  equal 
to  W.  The  general  effect  of  the  tension  T  is  then  to 
increase  the  thrust  on  the  upper  section  of  the  boom 
and  to  decrease  the  tension  in  the  tie. 

It  will  be  observed  that  the  values  which  have  been 
found  are,  in  general,  loads  and  not  stresses.  The 
stress  in  the  tie  is  pure  tension;  the  boom  acts  as  a 
column,  compressed  between  C  and  A  and  loaded  at 
C  and  F\  the  stresses  in  the  post  or  pillar  depend  on 


FORCE   AND   EQUILIBRIUM   POLYGONS  19 

the  method  of  connection  of  the  tie  and  boom  to  each 
other  and  to  the  vertical  member  between  them.  In 
the  ordinary  pillar  crane,  vertical  tie-rods,  connecting 
the  lower  ends  of  the  main  tie  and  boom,  take  the 
vertical  components  of  the  stresses  in  these  two  mem- 
bers. The  pillar  is  subjected  to  bending  by  the  force 
P%,  and  to  compression  from  the  direct  vertical  load 

Wy  and  there  is  both  a  vertical  reaction  equal  to  W  on 
the  pintle  at  the  upper  end  of  the  pillar  and  a  bending 
stress  due  to  Pz.  Hence,  to  determine  the  unit-stress 
in  the  metal,  the  conditions  under  which  each  member 
of  the  crane  works  must  be  considered. 

In  the  foregoing  the  weight  of  the  crane  has  been 
disregarded,  and  this  weight  must  be  taken  into  account 
for  an  accurate  determination  of  the  stresses.  The 
weight  of  any  member  of  any  structure  acts  at  the 
centre  of  gravity  of  that  member,  and  is  assumed  to  be 
divided  between  the  two  points  of  support  of  the  latter 
in  inverse  proportion  to  the  lengths  of  the  two  segments 
between  these  points  and  the  centre  of  gravity.  Thus, 
if  a  member  be  10  feet  long  and  weigh  1000  pounds, 
and  if  the  centre  of  gravity  be  7  feet  from  the  left  end, 
300  pounds  will  be  sustained  by  the  support  at  that 
end  and  700  pounds  by  the  right  support.  Proceeding 
thus,  we  have  a  series  of  vertical  forces,  due  to  this 
weight  or  '  dead  load,'  acting  at  such  joints  as  A,  B, 
and  C.  For  full  accuracy,  these  vertical  forces  must 
be  considered  in  drawing  the  force  polygons.  This 
method  is  general,  and  applies  to  all  structures.  If 

W^  be   the   total   weight   of   the   crane,   the   pressure 


2O 


GRAPHIC    STATICS 


on  the  bearing  at  the  base  of  the  pillar  will  be  W  -f  Wl 
=  Rf,  the  corresponding  vertical  reaction  there. 

The  weights  of  the  members  should  first  be  estimated, 
in  a  tentative  design,  from  the  data  of  cranes  previously 
built.  If  the  members  as  designed  exceed  these  weights, 
the  forces,  stresses,  and  dimensions  must  be  recalcu- 
lated until  a  reasonable  approximation  to  accuracy  is 
reached.  As  to  pillar  cranes  in  general,  a  reasonable 

estimate,  as  given 
by  good  practice, 
of  the  weight  of 
all  parts  except 
the  pillar  and  its 
immediate  connec- 
tions, is  one-half 
of  the  maximum 
'  live  load '  to  be 
carried  by  the  crane 
hook ;  the  distance 
of  the  centre  of 
f  gravity  of  these 
weights  from  the 
centre  of  the  pillar 
is  usually  about 
one-fourth  the  ra- 
dius or  swing  of 
the  crane. 

6.  SHEER  LEGS.  The  analysis  of  the  stresses  in 
sheer  legs  differs  from  the  preceding  examples,  since 
the  forces  and  stresses  are  not  all  in  the  same  plane. 


FIG.  10 


FORCE    AND   EQUILIBRIUM   POLYGONS  21 

The  legs  in  Fig.  10  are  130  feet  long;  the  angle  be- 
tween them  is  20  degrees ;  at  the  inclination  shown, 
the  plane  passing  through  them  makes  an  angle  of 
60  degrees  with  the  horizontal,  and  one  of  20  degrees 
with  the  tie  or  back  leg.  Taking  the  load  as  100  tons, 
it  is  required  to  find  the  loads  on  the  members.  The 
weight  of  the  structure  will  be  neglected. 

As  stated,  the  lines  of  action  of  the  loads  and  stresses 
lie  in  two  planes :  the  vertical  plane  passing  through 
the  tie  and  the  inclined  plane  in  which  the  legs  lie. 
As  graphic  methods  are  applied  to  forces  in  one  plane 
only,  we  must  find  first  the  resultant,  along  the  inter- 
section of  the  two  planes,  of  the  forces  in  the  inclined 
plane.  The 'joint  O  is  in  equilibrium  under  the  action 
of  this  resultant,  the  load  W,  and  the  stress  T  in  the 
tie.  In  the  force  triangle  (A)  lay  off  ab  —  W=  100 
tons ;  draw  be  and  ac  parallel,  respectively,  to  the  tie 
ON  and  the  legs  OM.  Then,  be  =  T  =  143  tons  is  the 
load  on  the  tie  or  back  leg,  and  ca  —  P  —  220  tons  is 
the  resultant,  along  the  intersection  of  the  two  planes, 
of  the  compressive  forces  on  the  sheer  legs. 

To  find  the  load  or  stress  Pl  on  each  leg,  the  re- 
sultant P  must  be  resolved  along  the  lines  of  the  legs. 
Therefore,  in  (B),  lay  off  ac  =  P,  and  from  a  and  c 
draw  ad  and  cd,  each  making  an  angle  of  2O°/2  =  10 
degrees  with  ac.  Then  cd  =  da  —  P1  =  112  tons.  The 
horizontal  component  of  da  or  P/2  is  ed  =  Q  =  22.5 
tons,  which  is  the  outward  force  acting  on  each  leg  at 
Mto  spread  the  pair  apart. 

At  the  joint  N  the  diagonal  tensile  force  T  is  resisted 


22  GRAPHIC   STATICS 

by  the  vertical  downward  pull  R  of  the  foundation  bolts, 
and  the  horizontal  tensile  stress  7\  of  the  screw  or 
other  mechanism  used  for  raising  or  lowering  the  legs. 
Hence,  joint  N  is  in  equilibrium  under  the  action  of 
the  forces  T,  R,  and  Tr  In  (C)  lay  off  cb  =  T\  draw 
bf  vertically  from  b ;  and,  horizontally  from  c,  draw  fc. 
Then  bf  =  R  =  91.5  tons,  andfc  =  7\  =  110.5  tons. 

In  these  diagrams  the  hoisting  rope  is  assumed  to  act 
along  the  tie;  if  its  line  of  action  be  inclined  to  the 
latter,  the  principles  of  Example  5  must  be  applied. 
With  screw  mechanism,  the  point  N  moves  to  and  fro 
horizontally  in  raising  or  lowering  the  she.er  legs ;  if  a 
tackle  be  substituted  for  the  tie  or  back  leg,  the  point 
N  should  be  fixed.  Each  of  the  sheer  legs  is  com- 
pressed like  a  column  by  the  force  Pl  and  its  weight; 
the  latter,  acting  at  the  centre  of  gravity  of  the  leg,  also 
produces  bending  stresses.  The  tie  is  in  tension,  but 
its  weight  acts  similarly  to  produce  bending.  Owing 
to  the  lengths  of  these  members,  their  weights  are 
relatively  large,  and  hence  the  effects  of  these  combined 
stresses  must  be  considered  carefully  in  designing  the 
parts. 


CHAPTER    II 

TRUSSES:  STRESS   DIAGRAMS 

5.  Framed  Structures.  The  term  *  framed  struc- 
tures '  is  used  to  designate  such  constructions  as  roof 
and  bridge  trusses,  the  lattice  girders  for  overhead 
cranes,  the  walking  beam  of  an  engine,  etc.  In  all  such 
structures,  ther  loads,  however  applied,  are  distributed 
among  the  members,  which  are  thus  simultaneously 
under  strain.  The  fundamental  requirement  in  truss 
design  is  that  the  members  shall  be  so  combined  as  to 
form  a  series  of  triangles  whose  sides  shall  be  capable 
of  withstanding  compressive  stress,  if  necessary.  This 
requirement  is  due  to  the  fact  that  a  triangle  —  assum- 
ing rigid  sides  and  hinged  vertices — is  the  only  poly- 
gon which  will  not  change  its  shape  when  loads  are 
applied  at  one  or  more  vertices.  The  members  form- 
ing a  truss  may  be  designed  for  tension  only  as  in  a  tie, 
for  compression  only  as  in  a  strut>  or  for  both  stresses 
as  in  the  tie-strut. 

(a)  Assumptions.  In  the  graphic  analysis  of  trusses, 
two  basic  assumptions  are  made  :  first,  the  loads,  of  what- 
ever character,  are  considered  as  transferred  to,  and  ap- 
plied at  the  joints  of  the  truss  only,  which  would  produce, 
only  direct  tension  or  compression  in  the  members 

23 


24  GRAPHIC    STATICS 

which  meet  at  a  joint;  and,  second,  these  loads  are  held 
in  equilibrium  at  each  joint  by  the  internal  stresses  act- 
ing in  the  members  united  there. 

In  practice,  the  first  of  these  conditions  is  met  by  so 
designing  the  structure  as  to  transfer  the  loads  to  the 
joints;  the  purlins  and  jack  rafters  of  a  roof  truss  and 
the  floor  system,  or  transverse  and  longitudinal  beams 
and  sleepers,  of  a  bridge  truss  serve  this  end.  It  is 
true  that  bending  stresses,  due  to  the  weight  of  the 
members,  and,  in  some  cases,  to  the  uniform  load  carried 
by  them,  do  exist  in  the  members,  and  when  the  latter 
are  relatively  large,  these  stresses  should  be  considered 
in  addition  to  the  direct  stresses  for  which  the  parts  are 
designed.  The  second  of  the  assumptions,  as  above, 
must  be  true  for  the  stability  of  the  structure. 

(b)  Loads.  All  trusses  carry  as  a  dead  load,  the  weight 
of  the  structure,  which  is  assumed,  in  general,  to  be 
divided  proportionately  and  concentrated  at  the  joints. 
In  roof  trusses  whose  roof-surface  is  inclined  at  less 
than  60  degrees,  there  is  in  winter  an  additional  uniform 
load  in  the  weight  of  the  snow  on  the  roof,  and  in  all 
such  trusses  there  is  also  an  intermittent  load,  in  the 
wind  pressure  on  one  side  or  the  other  of  the  roof. 
Bridge  trusses  have  similar  wind-loads,  but  their  chief 
burden  is  the  live  or  moving  load  which  crosses  the 
bridge.  These  statements,  as  to  dead  and  live  loads, 
apply  also  to  the  trusses  for  overhead  cranes. 

6.  Stress  Diagrams.  Since  it  is  assumed  that  all 
loads  are  concentrated  at  the  joints,  that  the  members 


TRUSSES:   STRESS  DIAGRAMS  25 

are  subjected  to  direct  axial  stress  only,  and  that  each 
joint  is  in  equilibrium  under  the  action  of  the  loads 
applied  there  and  the  stresses  whose  lines  of  action 
meet  at  that  point,  it  is  evident  that  the  joints  of  a  truss, 
such  as  is  shown  in  the  upper  diagram,  Fig.  1 1,  are  similar 
in  principle  to  the  joints  of  an  equilibrium  polygon.  It 
should  be  understood  that  the  truss  {force  or  space)  dia- 
gram is  not  such  a  polygon  and  that  this  similarity  re- 
lates only  to  the  joints  and  the  elementary  triangles  of 
the  truss. 

Owing  to  this  similarity,  there  can  be  drawn,  for  each 
joint,  a  force  polygon,  since  the  directions  of  all  lines  of 
action  and  the  magnitudes  of  the  external  loads  are 
known.  Further,  since  each  member  connects  two 
joints,  one  side  of  each  force  polygon  is  common  to  it 
and  to  the  polygon  immediately  following.  Hence,  the 
series  of  force  polygons,  for  a  complete  truss  or  other 
similar  structure,  is  consecutive  and  forms  the  stress 
diagram. 

The  character  of  the  stress  in  a  member — either  tensile 
( + )  or  compressive  (  —  )  —  is  indicated  by  the  direction 
of  the  stress  with  regard  to  the  joint,  being  tensile  if  it  act 
away  from  the  joint  and  compressive  if  it  press  toward 
the  latter.  This  direction,  to  or  from  the  joint,  is  given, 
as  shown  previously,  by  that  of  the  stress  in  passing 
around  the  force  polygon.  In  constructing  this  polygon, 
the  loads  and  stresses  about  a  joint  are  taken  usually 
in  clockwise  order,  and,  starting  at  any  joint,  the  remain- 
der may  be  taken  consecutively,  or,  as  is  sometimes 
necessary,  alternately  from  the  two  sides  of  the  truss. 


26 


GRAPHIC   STATICS 


The  stresses  in  the  members  of  a  truss  may  also  be 
determined  by  Rankine's  "  Method  of  Sections."  This 
is  an  analytical  process,  the  application  of  which  is 
shown  in  Art.  22. 

EXAMPLES 

7.  ROOF  TRUSS,  DEAD  LOAD.  Figure  1 1  gives  the 
skeleton  diagram  of  a  steel  roof  truss,  having  a  lower 

chord  A  A',  an 
upper  chord  com- 
posed of  rafters 
inclined  at  about 
30  degrees  to 
the  horizontal, 
and  connecting 
diagonal  mem- 
bers or  braces. 
The  span  A  A1 
is  36  feet,  the 
rise  from  chord 
to  peak  is  10 
feet,  each  rafter 
is  20  feet  9 
inches  long,  and 
the  trusses  are 
spaced  20  feet 
apart.  The 

joints     of     the 
FIG.  ii  J 

rafters  divide  the 

section  of  roof  carried  by  each  truss  into  panels  of  equal 
width,  1-2,  2-3,  etc. ;  the  length  of  the  panels  is  12  feet, 


TRUSSES:   STRESS  DIAGRAMS  27 

i.e.  the  distance  between  the  centre-lines  of  the  adjacent 
spaces  between  the  trusses.  The  total  weight  of  the 
truss  and  roof  covering  is  borne  by  the  rafters,  and  each 
joint  (apex  or  panel-point)  of  the  latter  is  assumed  to  sup- 
port one-half  of  the  load  on  each  of  the  adjacent  panels 
and  a  corresponding  share  of  the  weight  of  the  truss. 
In  Fig.  1 1,  there  are  thus  three  joints  having  a  full  panel- 
load,  and  one  at  each  of  the  two  supports,  A  and  A', 
having  one-half  of  that  load.  The  total  dead  load  is 
sustained  at  the  supports  by  the  two  vertical  reactions, 
R1  and  R^.  With  dead  and  snow  loads,  these  reactions 
are  equal  for  a  symmetrical  truss. 

The  dead  load  consists  of  the  weight  of  the  truss  and 
that  of  the  roof  covering.  In  designing,  the  weight  of 
the  truss  may  be  estimated  from  those  of  trusses  previ- 
ously built,  or  an  empirical  formula  may  be  used. 
Trautwine  gives,  for  preliminary  estimates  of  the 
weights  of  steel  trusses  in  pounds  per  square  foot  of 
building  space  covered,  0.05  to  0.09  x  span  in  feet. 
The  roof  covering  includes  the  jack  rafters,  purlins, 
sheathing,  and  the  outer  layer  of  felt  and  gravel,  or 
slate,  corrugated  iron,  etc.  The  weights,  per  square 
foot  of  roof  surface,  of  these  materials  are  given  in 
works  on  this  subject.  The  snow-load  varies  from  10 
to  30  pounds  per  square  foot  of  horizontal  projection 
of  the  roof;  it  is  about  20  pounds  in  the  latitude  of 
New  York  City ;  roofs  inclined  at  60  degrees  or  more 
have  no  snow-loads.  The  load  due  to  wind  pressure 
will  be  considered  in  Example  8. 

Figure  1 1  shows  one  of  the  methods  of  notation  used 


28  GRAPHIC    STATICS 

for  trusses.  Within  each  triangle  and  between  each 
pair  of  loads  or  forces,  capital  letters  are  placed,  and 
the  member  or  load  is  designated  by  the  letters  between 
which  it  lies.  Thus,  BD  is  the  lower  section  of  the  left 
rafter,  DE  the  adjacent  diagonal,  etc.  The  stresses  in 
these  members  are  indicated  in  the  force  polygon  by  the 
same  letters,  although  not  in  capitals.  Thus,  the  length 
of  the  side  Ogives  the  stress  in  the  member  BD,  etc.  The 
order  in  which  the  forces  about  a  joint  are  taken  in  con- 
'structing  the  force  polygon  is  frequently  shown  by  a  cir- 
cle with  arrowheads,  as  in  the  figure.  At  the  left  support 
A,  and  similarly  at  the  right  support,  there  are  two  op- 
posing forces  acting  vertically  :  the  half  panel-load  W/2 
and  the  total  reaction  Rl  of  the  support,  which  reaction  is 
equal  to  one-half  the  total  load  or  2  W.  The  difference, 
1.5  W,  between  these  two  forces  is  the  effective  reaction. 
Taking  13.5  pounds  as  the  weight,  per  square  foot  of 
roof  surface,  of  the  roof  covering  and  using  the  formula 
previously  given  for  the  truss  weight,  we  have,  for  the 
truss  shown  in  Fig.  1 1  : 

Area  of  roof  surface  =  12  x  20.75  x  2  —    49$  square  feet. 

Weight  of  roof  covering  =  498  x  13.5  =  6723  pounds 

Weight  of  truss  =  0.07  x  36  (12  x  36)  =  1244  pounds 

Dead  load  =  7967  pounds 

Total  reaction  =  R^  —  R.^  —  7967/2  =  3984  pounds 

Panel-load  =  W—  7967/4  =  1 992  pounds 

On  the  load-line  aal ',  lay  off,  to  any  convenient  scale 
of  pounds  to  the  inch,  the  loads  in  consecutive  order; 
thus,  a£  =  load  AB  =  W/2,  be  =  load  BC  =  W,  etc. 
Joint  I  at  the  left  support  is  in  equilibrium  under  the 


TRUSSES:   STRESS  DIAGRAMS 


29 


action  of  the  load  AB,  the  stresses  in  BD  and  DG,  and 
the  reaction  GA  or  R±.  In  the  force  polygon,  ab  is  the 
load  AB  and  ga  is  R1 ;  from  b  and  £•,  lay  off  bd  and  ^, 
parallel,  respectively,  to  BD  and  /?£.  The  closed  poly- 
gon for  the  joint  is  then  abdga,  and,  in  passing  around 
it,  bd  acts  toward  the  joint  and  is  compressive,  while  dg 
acts  from  the  joint  and  is  tensile. 

At  joint  2,  the  system  in  equilibrium  is  composed  of 
the  stress  DB  and  the  load  BC  which  are  known,  and 
the  stresses  in  CE  and  ED  which  are  to  be  .determined. 
From  c  and  d,  draw  ce  and  ed  parallel,  respectively,  to 
CE  and  ED  and  meeting  at  e.  The  closed  polygon  is 
then  bcedb,  and  the  character  and  magnitude  of  the 
stresses  can  be  found  as  before.  The  similar  polygon, 
similarly  constructed,  for  joint  3  is  cc'e'fec.  Since  both 
the  truss  and  the  system  of  loads  are  symmetrical,  the 
stresses  in  corresponding  members  on  the  two  sides  are 
the  same,  and  hence  the  stress  diagram  need  not  be 
considered  further. 

Tabulating  the  results,  as  measured  to  scale  from  the 
stress  diagram,  we  have : 


STRESS  DUE  TO  DEAD  LOAD 

Pounds 

Kind 

BD    

6080 



CE    ....... 

5080 

- 

ED    

1760 

- 

EF    

1760 

+ 

FG    

3640 

+ 

GD    

5280 

+ 

30  GRAPHIC    STATICS 

The  snow  load  is  virtually  an  additional  and  temporary 
layer  of  roof  covering.  Hence,  its  weight  produces  the 
same  kind  of  stress  in  each  member  as  the  dead  load, 
the  amount  thus  added  to  each  dead  load  stress  being 
proportional  to  the  ratio  of  the  snow  and  dead  loads. 
Taking  the  weight  of  the  snow  as  15  pounds  per  square 
foot  of  projected  area  of  the  section  of  roof  carried  by 
the  truss,  the  total  snow-load  is  36  x  12  x  15  =  6480 
pounds.  The  ratio,  as  above,  is  then  6480/7967  =  0.813, 
and  hence  the  total  stress  in  any  member,  due  to  the 
two  loads,  is  1.813  times  that  tabulated  above. 

8.  ROOF  TRUSS,  WIND  LOADS.  The  stresses  produced 
in  a  roof  truss  by  wind  pressure  are,  in  some  cases, 
opposite  in  kind  to  those  due  to  dead  and  snow  loads. 
Therefore,  a  separate  stress  diagram  is  drawn  for  wind 
loads,  and  the  results  measured  from  it  are  added  alge- 
braically to  the  dead  and  snow  load  stresses,  in  order  to 
obtain  the  final  stress  in  each  member.  The  latter  is 
designed  not  only  for  the  maximum  stress  of  either 
kind,  but  also  for  withstanding  stresses  of  both  kinds, 
if  the  wind  causes  a  reversal  of  stress. 

The  action  of  wind  pressure  on  an  inclined  roof  is 
not  understood  fully,  and  several  empirical  methods 
of  estimating  wind  loads  are  used.  Hutton's  formula 
is  : 


in  which  6  is  the  inclination  of  the  roof  surface  to  the 
horizontal,  pn  is  the  perpendicular  pressure  per  square 
foot  on  that  surface  and  due  to  the  wind,  and  ph  is  the 


TRUSSES:   STRESS  DIAGRAMS  31 

similar  pressure  on  a  vertical  plane  by  a  horizontal 
wind.* 

The  usual  practice  is  to  assume  ph  as  30  to  40  pounds. 
At  40  pounds  and  with  6  =  30  degrees,  pn  =  26  5 
pounds ;  and,  remembering  that  only  the  windward  side 
of  the  roof  is  subjected  to  wind  pressure,  we  have,  for 
the  truss  shown  in  Figs.  1 1  and  12  : 

Total  load  on  windward  side  =  20.75  x  12  x  26.5  =  6598.5  pounds 
Panel-load  on  windward  side  —  Wl  =  6598.5/2      =  3299  pounds 

Again,  metal  trusses  differ  in  their  methods  of  support. 
If  short,  both  ends  may  be  fixed;  if  long  and  large, 
one  end  is  fixed  and  the  other  is  free  to  allow  for 
expansion  and  contraction,  the  free  end  being  sup- 
ported on  a  roller  or  sliding  bearing.  If  the  truss  is 
fixed  at  both  ends,  but  one  stress  diagram  for  wind 
pressure  need  be  made,  with  the  wind  on  one  side  only, 
since,  if  the  wind  shift  to  the  other  side,  the  stresses  in 
the  two  halves  of  the  truss  will  simply  be  interchanged. 
When  one  end  of  the  truss  is  free,  however,  the  case  is 
different,  as,  disregarding  the  friction  of  the  bearing, 
the  reaction  at  the  free  end  is  always  vertical,  while  that 
at  the  fixed  end  varies  in  direction  and  magnitude  as  the 
wind  shifts. 

Let  it  be  required  to  find  the  stresses  due  to  wind 

*  A  simpler  expression  of  at  least  equal  practical  value  is  the  Straight 
Line  Formula: 

Pn  =  AP/4S 

in  which  A,  Pn,  and  P  correspond  with  0,  pn,  and  ph  in  Mutton's  formula. 
The  student  will  find  additional  expressions  for  the  normal  component  of  the 
wind  pressure  in  works  treating  this  branch  of  the  subject  in  greater  detail. 


32  GRAPHIC   STATICS 

loads  on  the  truss  whose  dead  load  stresses  were  de- 


FlG.    12 


termined  in  Example  7.     In  Fig.  12,  assuming  the  right 
end  of  the  truss  to  be  free  and  the  wind  to  be  on  the 


TRUSSES:   STRESS  DIAGRAMS  33 

left,  there  will  be  a  normal  load,  W-^  —  3299  pounds,  at 
joint  2,  and  a  similar  load  of  W^2  at  each  of  the  joints 
i  and  3.  As  the  dead  and  snow  loads  are  not  consid- 
ered, there  are  no  loads  at  joints  i'  and  2'.  The 
resultant  of  the  wind  loads  must  have  the  same  line  of 
action  as  the  middle  load,  since  this  load  is  the  greatest 
of  the  three  and  the  two  others  are  equal  and  equally 
spaced.  This  resultant  and  the  reactions  at  the  two 
supports  form  a  system  of  external  forces  in  equi- 
librium, and  hence  the  resultant  of  the  two  reactions 
must  be  equal  and  opposite  to  that  of  the  loads,  and 
have  the  same  line  of  action. 

As  the  right-hand  end  of  the  truss  is  free,  R%  will  act 
vertically.  The  lines  of  action  of  R%  and  the  load-resultant 
are  thus  known,  as  is  the  magnitude,  2  Wv  of  the  latter. 
Prolong  the  line  of  action  2(9  of  the  resultant  wind  load 
until  it  meets  at  O  that  of  R% ;  then,  since  the  forces  are 
concurrent,  Oi  is  the  line  of  action  of  Rr  On  2(7,  lay 
off  the  loads,  be  =  W^  and  ab  =  cc'  =  Wl/2.  From  a 
and  c',  draw  ag  and  c'g  parallel,  respectively,  to  Oi  and 
Oir.  The  force  polygon  for  the  loads  and  reactions  is 
then  ac'ga  and  ac'  =  loads,  c'g—  R2,  and ga  —  Rv 

Joint  i  is  in  equilibrium  under  the  action  of  Rl9  the 
load  Wl/2t  and  the  stresses  in  the  members  BD  and 
DG.  In  the  force  polygon,  ab  is  the  load ;  draw .  bd 
and  gd  parallel  to  BD  and  GD,  respectively.  The  force 
polygon  is  then  abdga,  and  bd  is  a  compressive,  and  dg 
a  tensile,  stress. 

At  joint  i',  the  system  in  equilibrium  is  the  reaction 
R<>  and  the  stresses  in  B' D1  and  D'  G.  As  there  are  no 


34  GRAPHIC    STATICS 

loads  at  joints  i'  and  2',  the  vertex  c'  in  the  force  poly- 
gon is  also  the  location  of  the  points  b1  and  a'.  Draw 
b'd'  and  gd1  parallel  to  the  two  members.  The  force 
polygon  is  then  a'gd'b'd ',  and  gd'  is  tensile  and  d'b' 
compressive. 

Proceeding  similarly  for  joint  2,  the  force  polygon  is 
bcedb.  In  the  polygon  for  joint  2',  b'd'  is  known  and  b' 
and  c'  lie  at  the  same  point.  Hence,  e'c'  must  coincide 
with  b' d' ,  and  the  force  polygon  is  b'd'e'c' b' ,  which  is  a 
straight  line.  There  is  no  stress  in  the  diagonal  D' E1 . 

In  the  polygon  at  joint  3,  there  are  known  the  three 
consecutive  sides,  ec,  ccl ',  and  c'e' ;  the  remaining  sides, 
e'faudfe,  must  meet  at  the  point/.  Hence, /coincides 
with  e',  and  there  is  no  stress  in  the  member  E1  F.  The 
polygon  is  cc'e'fec. 

Now,  let  the  wind  shift  to  the  right.  The  loads  (not 
shown)  due  to  it  will  be  removed  to  joints  i'  and  2',  and 
that  at  joint  3  will  be  parallel  to  them  and  normal  to  the 
right  rafter.  Their  resultant  will  pass  through  2'  and 
meet  the  new  right  reaction  R'%  —  which  is  vertical  as 
before  —  at  the  point  M' '.  Hence,  M'l  is  the  line  of 
action  of  the  new  left  reaction  R\,  which  line  practi- 
cally coincides  with  the  rafter,  as  the  inclination  of  the 
latter  is  about  30  degrees.  Prolonging  the  line  M'21 
for  the  load-line  a'c  and  proceeding  as  before,  the  force 
polygon  a'cga'  for  the  loads  and  reactions  is  drawn. 
The  stress  diagram  is  laid  out  by  the  previous  method, 
except  that  the  stresses  are  taken  in  non-clockwise  order. 

Tabulating  the  results,  as  measured  from  the  dia- 
grams in  Figs,  ii  and  12,  we  have: 


TRUSSES:   STRESS  DIAGRAMS 

FINAL   STRESSES 


35 


MEMBER 

DEAD 
LOAD 

SNOW 

1 

WIND  ON 

STRESS 

Fixed  Side 

Free  Side 

Maximum     Minimum 

BD        -  6080      -  4943      -  4600 

-  3760 

-  7-86    ,     -  3.04 

B'D'  \   -  6080      -  4943      -  3800 

~  4440 

-  7-73      -  3-04 

CE 

—  5080 

—  4130    :    —  4600 

-  3760 

-6.9 

-  2.54 

CE 

-  5080  j   -  4130 

-  3800 

—  4440    i    —  6.82 

-2.54 

DE 

—  1760      -  1431 

~3299 

o          -  3.24 

-0.88 

D'E 

-  1760      -  1431 

0 

-   3299         -  3.24 

-0.88 

EF 

+  1760  :  +  1431 

+  3200 

o          +3.19. 

+  0.88 

E'F 

+  1760  !  +  1431 

0 

+  3I2o 

+  3-16 

+  0.88 

DG 

+  5280 

+  4293 

-  +  6640 

0 

+  8.1 

+  2.64 

D'G 

+  5280 

+  4293 

+  3260 

+  2920 

+  6.41 

+  2.64 

FG 

+  3640 

+  2959 

+  3260 

0 

+  4-93 

+  1.82 

The  maximum  and  minimum  stresses  are  given  in 
tons  (2000  pounds),  the  remainder  in  pounds.  The 
maximum  stress  is,  in  each  case,  the  algebraic  sum  of 
the  dead  and  snow  loads  and  the  greater  of  the  two 
wind  loads ;  the  minimum  is  the  dead  load  stress.  As 
before,  tensile  stresses  are  marked  -f  and  compressive 
stresses  — . 

9.  CRANE  TRUSS.  Trusses  are  used  frequently  in 
crane  construction.  Thus,  the  crane  shown  in  Fig.  13 
is  virtually  a  braced  cantilever,  the  vertical  and  diagonal 
members  serving  to  stiffen  the  structure  and  to  dis- 
tribute the  dead  load,  which  is  the  weight  of  the  crane. 
The  live  load  W19  suspended  from  the  peak  A ',  is,  from 
the  peak  to  the  diagonal  PQ,  assumed  to  be  taken  solely 
by  the  upper  and  lower  chords  or  sides  of  the  boom,  the 


GRAPHIC   STATICS 


FIG.  13 


TRUSSES:   STRESS  DIAGRAMS  37 

section  between  joints  i  and  10  acting  as  a  tie  and  that 
from  I  to  9  as  a  strut.  This  assumption  is  sufficiently 
correct  for  general  purposes,  but,  for  an  absolutely 
accurate  determination  of  the  stresses  in  the  members, 
the  effect  of  bending  should  be  considered  also. 

Let  the  total  weight  of  the  crane  or  dead  load  be 
1800  pounds  and  the  maximum  live  load  or  weight  to 
be  carried  at  the  peak  be  W^  —  5  tons  (10,000  pounds). 
Assume  the  dead  load  to  be  distributed  among  the 
upper  joints,  as  in  a  roof  truss,  the  panel-load  W  being 
thus  300  pounds.  While  this  distribution  is  not  strictly 
that  which  actually  exists,  it  is  generally  used,  for 
cranes  of  moderate  size,  as  a  ready  and  reasonable 
approximation. 

A  single  stress  diagram  could  be  drawn  for  both  the 
dead  and  live  loads,  if  the  load  at  joint  i  were  taken  as 
the  total  load  there,  or  W1  +  W/2 ;  but  Wl  is  so  much 
greater  than  W  that  the  scale  necessary  for  accurate 
measurement  of  the  stresses  in  the  members  would 
make  the  diagram  inconveniently  large.  Hence,  sepa- 
rate diagrams,  at  different  scales,  will  be  drawn  for  the 
dead  and  live  loads.  The  total  stress  in  a  member  will 
then  be  the  algebraic  sum  of  those  given  by  the  two 
diagrams. 

In  the  dead  load  diagram,  lay  off  the  loads  from 
a'  to  77,  taking  them  in  anti-clockwise  order.  As  W1 
is  disregarded,  the  point  a'  corresponds  also  with 
the  point  /.  Joint  I  is  in  equilibrium  under  the 
action  of  the  load  A' A  and  the  stresses  in  the  members 
AG  and  GT  —  GA.'.  Drawing  the  stress  lines  parallel 


38  GRAPHIC    STATICS 

to  their  corresponding  members,  we  have  the  force 
triangle  aga ',  in  which  ag  is  a  tensile,  and^;  =  gt  is  a 
compressive  stress. 

At  joint  2,  the  known  forces  are  GA  and  AB  =  W\ 
their  resultant  is  gb.  From  g  and  b,  lay  off  gh  and  bh 
parallel,  respectively,  to  HG  and  BH  and  meeting  at  h. 
The  force  polygon  is  then  gabhg.  The  brace  HG  thus 
acts  as  a  strut,  which  is  the  case  with  the  similar  braces 
parallel  to  it. 

The  known  stresses  at  joint  3  are  a1  g  and  gh  ;  their 
resultant  is  a1  h.  From  a1  and  k,  lay  off  afk  and  hk 
meeting  at  k.  The  force  polygon  is  then  a'ghkct ,  and 
HK  is  a  tie,  as  are  LM  and  7W7.  The  polygons  for 
joints  4  to  8  are  drawn  similarly. 

At  joint  9,  the  known  stresses  are  a'c  and  op ;  their 
resultant  is  a'p.  Lay  off  afq  and  pq  parallel  to  their 
corresponding  members.  The  force  polygon  is  a'opqa!-, 
and  PQ  is  a  strut. 

The  force  polygon  for  joint  10  is  qpefsq,  and  SQ  is  a 
strut.  At  joint  11,  the  polygon  is  a! qsua!  ;  S£/  is  a 
strut,  and  ua'=R2,  the  right  and  upward  reaction  from 
the  foundation.  The  similar  polygon  for  joint  12  is 
Msfvu,  and  vu  =  R^  is  the  downward  pull  of  the  founda- 
tion bolts. 

Now,  disregard  the  weight  of  the  crane  and  consider 
only  the  effects  of  the  live  load  W^  at  the  peak.  The 
general  method  and  the  order  in  which  the  stresses  are 
taken  are  the  same  as  before.  In  the  live  load  dia- 
gram lay  off  Wl  from  t  to  a'.  Joint  I  is  in  equilib- 
rium under  the  action  of  Wl  and  the  stresses  in  the 


TRUSSES:   STRESS  DIAGRAMS  39 

members  AG  and  GT.  As  there  are  no  dead  loads  at 
the  intervening  joints,  each  of  these  stresses  is  uniform 
in  intensity  throughout  the  members  from  joint  I  to 
joints  10  and  9,  respectively,  and  therefore  the  point  a' 
coincides  with  the  points  a,  b,  c,  d,  e,  and/,  and  simi- 
larly the  point  g  corresponds  with  the  points  k,  m,  o, 
and /.  From  t  and  a,  lay  off  tg  and  agy  parallel  to  GT 
and  AG,  respectively,  and  meeting  at  g.  The  force 
polygon  is  then  tagt\  ag  is  the  tensile  stress  in  the 
upper  members  between  joints  I  and  10;  gtis  the  stress 
in  the  members  forming  a  strut  between  joints  I  and  9 ; 
and  there  is  no  stress,  except  that  due  to  bending,  in  the 
diagonal  braces  from  joint  I  to,  and  including,  the 
member  OP. 

At  joint  9,  the  stresses  acting  are  those  in  the  mem- 
bers TO,  PQ,  and  QT,  that  in  OP  being  zero.  The 
force  polygon  is  topqt,  and  PQ  is  a  strut.  The  similar 
polygon  for  joint  10  is  qpefsq,  in  which  the  stress pe=ga 
and  ef  is  zero.  FS  and  SQ  are  under  tension.  For 
joint  n,  the  polygon  is  tqsut\  5£/acts  as  a  strut  and  ut 
gives  the  magnitude  of  R2,  the  right  reaction  and  up- 
ward pressure  from  the  foundations.  Finally,  the  poly- 
gon for  joint  12  is  usfvu,  in  which  the  point  v  coincides 
with/,  and  vu  =  Rl  is  the  downward  pull  of  the  founda- 
tion bolts. 

Tabulating  the  stresses,  as  measured  from  the  two 
diagrams,  and  taking  the  algebraic  sum  to  obtain  the 
total  stress  in  each  member,  we  have : 


GRAPHIC    STATICS 


MEMBER 

AG 
BH 
CL 
DN 
EP 
FS 

STRESS  (Ls.) 

DUE  TO 

TOTAL 
STRESS 

MEMBER 

STRESS  (Ls  ) 

DUE  TO 

TOTAL 
STRESS 

Dead 
Load 

Live 
Load 

Dead 
Load 

Live 
Load 

+    940 
+    820 
+  1780 
+  2640 

+  3540 
+  2170 

+  65,000 
+  65,000 
+  65,000 
+  65,000 
+  65,000 
+  34,000 

+  65,940 
+  65,820 
+  66,780 
+  67,640 
+  68,540 
+  36,170 

GH 
KL 

MN 
OP 

QS 

—  280 
—  400 

—  530 
-660 
—  480 

o 
o 
o 

0 

+  3500 

—    280 
-    400 
-    530 
-    660 
+  3020 

HK 
LM 

NO 
PQ 
SU 

+    940 

+  1010 
+  1100 

—  1640 
—  1480 

o              +   940 

0                   +  1010 
0                   +  IIOO 

—  37,000  i  —  38,640 
—  23,500  !  -  24,980 

GT 
KT 

MT 

or 

QT 

—  1000 

—  2080 
—  3060 
—  4080 
—  3240 

—  69,500 
—  69,500 
—  69,500 
—  69,500 
—  44,500 

—  70,500 
-71,580 
—  72,560 
-  73.580 
—  47,740 

Similarly,  the  reactions  are  : 


DEAD  LOAD 

LIVE  LOAD 

TOTAL,  POUNDS 

Rl 

145° 

24,000 

25.450 

#2 

325° 

34,000 

37.250 

CHAPTER    III 

STATIONARY   LOADS:    SHEARS   AND    MOMENTS 

7.  Beams.  In  the  analysis  of  their  stresses,  many 
machine  and  structural  members  are  classed  as  '  beams,' 
since  the  effects  of  their  applied  loads  are  the  same  in 
character  as  those  produced  by  the  loads  on  a  beam. 
Technically,  a  beam  is  a  rigid  bar,  generally  horizontal, 
and  sustained  by  one  or  more  supports.  A  simple  beam 
rests  freely  on  two  supports,  one  at  each  end ;  a  con- 
tinuous beam  has  three  or  more  supports ;  a  cantilever 
beam  has  but  one  support,  which  is  at  one  end,  this 
definition  including  the  part  of  any  other  form  of  beam 
which  projects  beyond  the  support  which  upholds  it. 

Supports  may  be  either  free  or  restrained.  With  the 
former,  the  beam  simply  lies  on  its  supports ;  when  the 
beam  is  restrained,  its  end  or  ends  are  built  in  or  other- 
wise so  fixed  that  the  tangent  to  the  elastic  curve  at  the 
supports  is  horizontal.  The  elastic  curve  is  the  curve 
into  which  the  neutral  axis  or  neutral  surface  of  the 
beam  is  bent  by  the  loads.  This  bending  produces  com- 
pression on  one  side  of  the  beam  and  tension  on  the 
other ;  the  neutral  axis  lies  in  the  plane  of  zero  stress 
between  the  compressed  and  stretched  parts  and  passes 
through  the  centre  of  gravity  of  the  cross-section  of  the 
beam. 

41 


42  GRAPHIC   STATICS 

8.  Vertical  Shear.  The  principal  effects  of  the  loads 
on  a  beam  are  to  produce  vertical  shear  and  bending 
moments.  Thus,  Fig.  14  (A)  represents  a  simple  beam 
—  ~~  _  ___^  without  loads.  When 

J       under    load,  it  is  sub- 
jected    simultaneously 

m)f~  to  shearing,  as  in  (B\ 

and  to  bending,  as  in 

(0- 

At  any  section,  as 
ab,  it  will  be  seen  that, 
in  shearing,  the  seg- 
ment of  the  beam  to  the  left  of  ab  is  forced  vertically 
upward  and  that  to  the  right  vertically  downward  by 
two  equal  and  opposite  forces  acting  at  the  section, 
each  known  as  the  vertical  shear  V.  The  shear  to  the 
left  is  that  usually  considered  ;  it  is  equal  to  the  result- 
ant or  algebraic  sum  of  the  forces  to  the  left  of  any 
section  as  ab,  upward  forces  and  shears  being  consid- 
ered as  positive,  and  those  acting  downward  negative. 


Thus  :  V=Ri 

in  which  Rl  is  the  left  reaction  and  2<P  is  the  algebraic 
sum  of  the  upward  and  downward  forces  between 
the  section  and  the  left  support.  In  (&),  if  the 
weight  of  the  segment  of  the  beam  to  the  left  is 
neglected,  ^P  =  P1  ;  if  this  weight  is  considered, 
2P  =  P1  +  wx,  in  which  w  is  the  weight  of  the  beam 
per  unit  of  length  and  x  is  the  distance  of  the  section 
ab  from  the  left  support. 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          43 

9.  Bending  Moment.     Similarly,  at  any  section,  as  ab 
in  (C\  Fig.  14,  the  forces  to  the  left  of  the  section  com- 
bine to  rotate  the  left  segment  in  a  clockwise  direction, 
while  those  to  the  right  act,  with  an  equal  resultant  bend- 
ingmoment^  to  produce  rotation  in  the  reverse  direction. 
The  results  are  to  compress  the  upper  side  of  the  beam, 
stretch  the  lower  side,  and,  if  the  load  be  excessive, 
to  rupture  the  fibres  at  the  bottom,  and  ultimately  the 
beam.     These  actions  are  the  effects  of  the  moments, 
about  any  point  in  the  section  ab,  of  the  forces  acting 
to  the  left  or  right  of  that  section,  the  resultant  bending 
moment  of  the  forces  to  the  right  being  equal  to  that  of 
those  on  the  left. 

Since  the  bending  moment  is  the  product  of  a  force 
by  its  distance  from  the  section,  it  is  a  compound  quan- 
tity, expressed  in  pounds-inches  or  tons-feet.  Neglect- 
ing the  weight  of  the  beam,  the  bending  moment  at  the 
section  ab  is  the  algebraic  sum  of  the  moments  of  the 
forces  to  the  left  of  that  section,  or : 

M=  R1  xx  —  Pl  x  xv 

The  weight  of  the  segment  to  the  left  will  act  at  the 
centre  of  gravity  of  that  segment.  Considering  this 
weight,  the  bending  moment  becomes : 

M  —  Rl  x  x  —  Pl  x  &\  —  ivx  X  xJ2. 

10.  Resisting  Moment.     The  resisting  moment  at  any 
section  of  a  beam  is  the  algebraic  sum  of  the  moments 
of  the  internal  horizontal  stresses  about  any  point  in  the 
section  considered,  these  stresses  being  produced  by  the 


44 


GRAPHIC    STATICS 


bending  moment.  For  equilibrium,  the  bending  moment 
and  resisting  moment  must  be  equal.  Hence,  as  is 
shown  in  works  on  Strength  of  Materials : 

Resisting  Moment  =  SI jc  =  Bending  Moment  =  M, 
in  which  S  is  the  unit-stress  on  the  fibre  most  remote 
from  the  neutral  axis  of  the  cross-section  of  the  beam  at 
the  section  considered,  c  is  the  distance  of  that  fibre 
from  the  neutral  axis,  and  /  is  the  rectangular  moment 
of  inertia  of  the  cross-section. 


11.  Shear  and  Moment  Diagrams.  The  vertical  shear 
and  bending  moment  at  any  section  of  a  beam  can  be 
found  from  the  force  and  equilibrium  polygons.  Thus, 


let  AB,  Fig."  15,  represent  a  simple  beam,  having  three 
vertical  loads  of  100,  400,  and  250  pounds,  spaced  as 
shown.  Lay  off  these  loads  on  the  load-line  ad,  draw 


STATIONARY    LOADS:     SHEARS    AND    MOMENTS          45 

the  rays  Oa,  Ob,  Oc,  and  Od,  and  construct  the  corre- 
sponding equilibrium  polygon  CDEFGC,  the  closing 
side  GC  of  which  determines  the  magnitude  ea  of  the 
left  reaction  R1  and  the  similar  magnitude  de  of  R2. 

The  vertical  shear  to  the  left  of  any  section  of  a  beam 
is,  by  definition,  equal  to  the  left  reaction,  minus  the 
sum  of  the  loads  to  the  left  of  that  section.  Hence, 
between  the  left  support  and  the  load  Pv  F=  Rl  =  ea ; 
between  P1  and  P2,  V— R^  — P^  —  ea  —  ab  —  eb;  between 
PI  and  /»8,  V=Rl-(Pl  +  Pi)  =  ea-ac=-ec\  and,  be- 
tween PB  and  the  right  support,  1/=Rl  —  (Pl-i-P2  +  PB) 
=  ea  —  ad=  —  ed=  —  R^.  Projecting  these  values  from 
the  load-line  on  the  lines  of  action  of  the  forces,  we 
have  the  vertical  shear  diagram  efkgh.  The  zero-line 
of  this  diagram  is  eg,  the  shear  above  it  being  positive 
or  upward,  and  that  below  it,  negative  and  downward. 

Now,  let  it  be  required  to  find  the  bending  moment 
at  any  section,  as  T,  of  the  beam,  the  corresponding 
ordinate  of  the  equilibrium  polygon  being  tt' .  The 
magnitude  of  the  bending  moment  depends  on  the 
magnitudes  and  locations  of  the  forces  to  the  left  of  T\ 
these  forces  are  P1  and  Rv  whose  resultant  r  is  equal  to 
eb  and  acts  vertically  upward  through  a  point  yet  to 
be  determined. 

Assume  the  polygon  to  be  cut  on  the  line  tt' .  Then, 
to  maintain  equilibrium,  the  stresses  5  and  S'  in  the 
sides  Ct1  and  Dt,  respectively,  must  have  a  resultant 
which  is  equal  and  opposed  to  r  and  which  has  the  same 
line  of  action.  The  resultant  r  and  these  two  stresses 
thus  form  a  system  of  concurrent  forces  in  equilibrium, 


46  GRAPHIC   STATICS 

and  hence  their  lines  of  action  must  meet  at  the  com- 
mon point  m,  the  intersection  of  t'C  and  tD,  prolonged. 
The  line  of  action  of  r  therefore  passes  through  the 
point  m,  distant  /  from  T,  and  the  bending  moment  M 
at  the  section  T  is  r  x  /. 

Let  H  be  the  pole-distance  in  the  force  polygon.     The 
triangles  mtt'  and  Oeb  are  similar.     Hence  : 

eb  \tf\\Oe\  mt 
r:tt':\  H\l 


That  is,  as  H  is  the  same  for  all  the  forces  and  as  the 
ordinate  tt'  is  parallel  to  the  lines  of  action  of  the  latter, 
the  bending  moment  at  any  section  of  the  beam  is  propor- 
tional to  the  corresponding  ordinate  of  the  equilibrium 
polygon,  and  is  equal  to  the  product  of  the  length  of  that 
ordinate  by  the  pole-distance.  This  principle  holds  only 
when  the  lines  of  action  of  the  forces  acting  on  the 
beam  are  parallel  and  the  section  considered  is  parallel 
to  those  lines. 

Under  these  conditions,  the  bending  moment  diagram 
is  the  equilibrium  polygon  for  the  forces  acting  on  the 
beam.  Figure  15  represents  this  diagram  for  concen- 
trated loads.  If,  as  in  Fig.  16,  the  beam  is  covered  by 
a  uniform  load  of  w  pounds  per  unit  of  length  and  the 
span  be  s,  the  left  reaction  will  be  one-half  of  the  total 
load,  or  ws/2t  and,  at  any  section,  distant  x  from  the 
left  support,  the  bending  moment  will  be  : 

M  =  wsxJ2  —  wx  x  xJ2 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          47 

which  is  the  equation  of  a  parabola  whose  maximum 
ordinate  ws2/S  occurs  at  the  middle  of  the  beam,  where 
x  =  s/2. 

As  with  the  vertical  shear,  the  bending  moment  may 
be  either  positive  or  negative.  For  positive  moments, 
the  diagram  is  usually  drawn  above  the  zero-line,  and, 
for  negative  moments,  below  it,  although  this  practice  is 
not  invariable. 

12  Moment  Scale.  To  make  practical  use  of  the 
equilibrium  polygon  as  a  bending  moment  diagram,  the 
scales  of  the  force  and  equilibrium  polygons  must  be 
considered.  Thus  if,  in  the  equilibrium  polygon,  Fig. 
15,  the  spacing  of  the  forces  be  on  a  linear  scale  of  4  feet 
to  the  inch,  or  48  to  I,  the  length  of  the  ordinate  tt\  as 
measured  in  inches  from  the  diagram,  must  be  multi- 
plied by  48  to  give  the  result  in  pound-inches,  or  by  4 
to  obtain  it  in  pound-feet.  Again,  if  the  force  polygon 
be  laid  out  on  a  force  scale  of  400  pounds  to  the  inch, 
and  the  measured  length  of  the  pole  distance  H  be  \\ 
inches,  then  H=  1.25  x  400=  500  pounds.  The  mo- 
ment scale,  by  which  tt'  and  similar  ordinates  are  to  be 
measured,  is  then  the  product  of  the  linear  scale  by  the 
pole  distance  measured  on  the  force-scale  or,  in  this  case  : 

4  x  500=  2000  pound-feet  per  inch, 
to  obtain  M  in  pound-feet,  or 

48  x  500  =  24,000  pound-inches, 
to  obtain  Mm  pound-inches. 


4  GRAPHIC    STATICS 

On  the  linear  scale,  as  above,  the  measured  length  of 
//  in  the  original  drawing  was  0.725  inch,  and  the 
bending  moment  M  at  the  section  T  is  therefore 

0.725  x  2000=  1450  pound-feet, 
or  0.725  x  24,000  =  17,400  pound-inches. 

13.  Twisting  Moments.  Shafting  transmits  power  by 
torsion.  The  driving  pulley  pulls  forward,  the  driven 
pulley  drags  backward,  and  hence  the  shaft  between 
them  is  twisted  by  an  amount  depending  on  its  material 
and  the  resistance  to  its  forward  motion.  For  equilib- 
rium, the  moment  of  the  driving  force  must  be  equal 
to  that  of  the  resistance,  and  therefore  the  twisting 
moment  is  uniform  throughout  the  shaft  between  the 
driving  and  driven  pulleys. 

The  effect  of  torsion  is  to  produce  shearing  stress  on 
the  shaft  in  planes  perpendicular  to  the  axis.  The 
moment  of  resistance  of  the  shaft  to  twisting  is  S8J/c, 
which  is  evidently  equal  to  the  driving  moment.  Sg  'is 
here  the  unit  shearing  stress  at  the  distance  c  of  the 
most  remote  fibre  from  the  axis,  i.e.  at  the  outer  extrem- 
ity of  the  radius,  and  J  is  the  polar  moment  of  inertia 
of  the  cross-section  of  the  shaft.  For  a  solid  cylin- 
drical shaft  of  diameter  d^J—  Trd^/^2 ;  for  a  hollow 
cylindrical  shaft  of  external  diameter  d-^  and  internal 
diameter  </,/ =  -rr(d£  -  ^4)/32. 

The  driving  moment  —  and  hence  its  equal,  the  twist- 
ing moment  —  may  be  expressed,  as  with  bending  mo- 
ments, in  pound-inches,  pounds-  or  ton-feet,  etc.  Since 
the  twisting  moment  is  uniform  throughout  the  length 


STATIONARY    LOADS  :     SHEARS   AND    MOMENTS          49 

of  shafting  under  torsion,  its  diagram  is  simply  a  rect- 
angle of  that  length  and  having  a  height  equal  to  the 
moment  expressed  in  pound-inches,  etc.,  to  any  given 
moment-scale,  as  shown  in  Fig.  19. 

14.  Bending  and  Twisting  Moments  combined  :  Equiva- 
lent Bending  and  Twisting  Moments.  Shafts  may  be 
subjected  to  both  twisting  by  the  driving  moment  and 
to  bending  by  the  weights  of  gears  and  pulleys,  the 
pull  of  belt-tensions,  etc.  In  such  cases,  in  place  of 
finding  the  stress  due  to  each  of  these  actions  separately 
and  taking  their  sum,  it  is  more  convenient  to  compute 
the  stress  corresponding  with  an  assumed  bending  or 
twisting  moment  which  is  equivalent  in  its  effect  to 
the  original  bending  moment  and  the  twisting  moment 
combined. 

Let  Mb  be  the  bending  moment,  Mt  the  twisting  mo- 
ment, and  EMb  the  equivalent  bending  moment.  Then, 
it  can  be  shown,  by  the  relation  between  the  polar  and 
rectangular  moments  of  inertia  for  a  circular  cross-sec- 
tion, that  : 

EMb  =  \(Mh 


Similarly,  letting  EMt  be  the  equivalent  twisting  mo- 
ment, we  have  : 


EM,  =  Mb  +  VMb*  +  M?. 

These  expressions  hold  only  for  solid  cylindrical  bodies, 
like  a  shaft. 

Bending  moments,  when  laid  out  by  the  same  moment 
scale,  can  be  treated  exactly  like  forces,  i.e.,  a  line  can 


50  GRAPHIC   STATICS 

be  drawn  representing  each  moment,  and  a  force  poly- 
gon constructed  for  a  series  of  these  moments.  The 
resultant  bending  moment  of  two  or  more  such  mo- 
ments in  different  planes  can  thus  be  found.  Simi- 
larly, twisting  and  bending  moments,  drawn  to  the  same 
moment  scale,  can  be  combined  to  form  the  equivalent 
bending  or  twisting  moment  by  applying  the  equations 
just  given.  Example  13  shows  the  methods  for  these 
combinations. 

EXAMPLES 

10.  /-BEAM:  UNIFORMLY  DISTRIBUTED  LOAD.  —  The 
methods  of  treating  a  uniformly  distributed  load  are  an 
extension  of  those  used  for  concentrated  loads,  as  shown 
in  Fig.  15.  When  the  load  is  uniform  over  the  entire 
span  of  a  beam,  the  number  of  equal  loads  or  parallel 
forces  becomes  indefinitely  large,  since  the  weight  of 
each  elementary  particle  of  the  load  acts  at  the  centre 
of  gravity  of  that  particle.  In  finding  the  bending 
moment  at  any  section  of  the  beam,  these  elementary 
forces  to  the  left  of  that  section  may  be  replaced  by 
their  resultant,  which  acts  at  the  centre  of  gravity  of 
the  total  load  to  the  left  and  is  equal  in  magnitude  to 
that  load. 

Thus,  in  Fig.  16,  let  AB  be  a  simple  beam  of  span  s 
and  carrying  a  uniformly  distributed  load  of  w  pounds 
per  unit  of  length.  The  total  load  is  then  w  x  s  =  W 
the  left  reaction  is  W/2  —  ws/2,  and  the  bending  mo- 
ment at  any  section  a,  distant  x  from  the  left  support, 
is  (Art.  9) : 

M=  WS/2  XX  —  WX  X  X/2, 


STATIONARY   LOADS:    SHEARS   AND   MOMENTS 


since  wx  is  the  total  weight  of  that  portion  of  the  load 
between  a  and  A,  x/2  is  the  distance  of  the  centre  of 
gravity  of  wx  from  a,  and  hence  the  lever-arm  of  that 
weight,  and  x  is  the  simi- 
lar arm  of  the  left  reaction 
Rv     This  expression   for   A 
M  is   the   equation    of   a 
parabola,  as  A'a'B1 ',  whose 
ordinate  is   zero    at   each 
support  and    a   maximum 
at  mid-span.     At  the  latter 
point,   x  —  s/2    and    M= 
ws*/8  =  Ws/8. 

The  curve  of  bending 
moments  can  be  drawn  by 
computing  the  value  of 
the  moment  at  mid-span, 
laying  off  this  value  there 
to  any  suitable  moment  scale,  and  then  describing  a 
parabola  through  the  point  D1  thus  obtained,  and  the 
points  A'  and  B1.  If  the  total  load  be  divided  into  any 
number,  as  three  equal  parts,  vertical  lines  of  action  be 
drawn  through  the  centres  of  gravity  of  these  parts,  the 
three  equal  loads  be  set  off  on  the  load-line  be,  and  the 
force  and  equilibrium  polygons,  Obc  and  A'C D'E'B', 
be  constructed,  it  will  be  found  that  the  points  Cr,  D' , 
and  E'  will  lie  on  the  parabolic  curve.  If  the  load  be 
divided  into  an  indefinitely  large  number  of  equal  parts, 
the  equilibrium  polygon  and  the  curve  will  coincide. 

As  an  example,  take  a  1 5-inch  /-beam  of  16  feet  span, 


FIG.  16 


52  GRAPHIC    STATICS 

weighing  50  pounds  per  foot  of  length,  and  carrying  a 
uniformly  distributed  load  of  43,000  pounds,  including 
its  own  weight.  It  is  required  to  determine  the  maxi- 
mum fibre  stress  in  pounds  per  square  inch. 

The  maximum  stress  will  occur  at  mid-span,  where 
M  =  Ws/8.  By  the  common  theory  of  flexure,  M  is 
also  equal  to  SI  fc  (Art.  10).  Equating  : 

Sf/c=  Ws/8  =  M, 

in  which  S  is  the  maximum  stress,  /  is  the  rectangular 
moment  of  inertia  of  the  cross-section  of  the  beam  about 
a  neutral  axis  passing  through  the  centre  of  gravity 
of  that  cross  section  and  perpendicular  to  the  web, 
c  =  15/2  =  7.5  is  the  distance  of  the  most  remote  fibre 
from  the  neutral  axis,  ^=43,000  pounds,  and  s—  16 
feet  =  192  inches.  For  this  beam,  1 '  —  483.4,  and  hence 
I/c  =  64. 5 .  Substituting : 

5  x  64.5  =  43,000  x  192/8 

5  =  16,000  pounds  per  square  inch. 

The  maximum  bending  moment  is 

5  x  64.5  =  1,032,000  pound-inches, 

which  is  the  moment  given  by  the  ordinate  at  D1 1  Fig. 
1 6,  when  that  ordinate  is  measured  by  the  moment 
scale  computed  as  described  in  Art.  12. 

ii.  LOCOMOTIVE  SIDE  ROD  WITH  UNIFORM  LOAD 
DUE  TO  CENTRIFUGAL  FORCE.  The  parallel  or  side 
rod  connecting  the  two  driving  wheels  of  a  locomotive, 
is  an  example  of  a  simple  beam  carrying  a  uniformly 
distributed  load  produced  by  centrifugal  force.  Every 


STATIONARY   LOADS:     SHEARS   AND    MOMENTS 


53 


point  in  the  rod  is  revolving  in  a  circle  whose  radius  is 
the  length  of  the  crank.  When  a  body  of  weight  W 
rotates  about  an  axis  in  a  circle  of  radius  r  with  a  uni- 
form velocity  v,  the  centrifugal  force  acting  radially 
outward  on  the  body  is: 


in  which  g  is  the  acceleration  of  gravity,  or  32.2  feet 
per  second  per  second. 


21     t 

F 


FIG.  17 

On  each  elementary  particle  of  the  revolving  side 
rod,  there  is  acting,  therefore,  a  centrifugal  force  parallel 
to  the  crank.  These  forces  have  their  maximum  effect 
in  bending  the  rod,  when  the  latter  is  in  its  highest  or 
lowest  position,  as  shown  in  Fig.  17,  and  the  lines  of 
action  of  the  forces  are  perpendicular  to  the  axis  of  the 
rod ;  when  the  rod  is  on  the  line  of  centres,  the  forces 


54  GRAPHIC    STATICS 

act  along  its  axis  to  produce  direct  tension  or  compres- 
sion ;  at  any  intermediate  position,  their  effect  on  the 
rod  is  partly  flexural  and  partly  to  produce  direct  stress, 
as  above. 

Let  V  be  the  circumferential  velocity  of  the  driving 
wheels,  which  is  that  of  the  train  ;  v,  the  corresponding 
velocity  of  the  crank  pins  ;  and  R  and  r,  the  radii  of  the 
driving  wheels  and  the  crank  pin  circles,  respectively. 
Then: 

v:  V\\r\R\ 

Vr 
V  =  -R' 

Again,  let  w  be  the  weight  of  a  unit  of  length  of  a 
side  rod  of  length  /,  and  let/  be  the  corresponding  cen- 
trifugal load  when  the  rod  is  in  its  highest  or  lowest 
position  and  the  bending  effect  is  a  maximum.  Then 
(Example  10),  the  maximum  bending  moment  is 

M=fl*/S  =  Ft/S  =  SI/c. 
But 

f=  wi^/gr  — 
Substituting, 


and 

5  =  Mc/I=(wrV*PI*gFF)  x  c/I, 

which  are  the  expressions  for  the  bending  moment  and 
maximum  stress  at  the  mid-length  of  the  rod. 

In  the  force  polygon,  Fig.  17,  let  ab  =  F=fl=\he, 
total  centrifugal  load.  Construct  the  force  polygon 
Oab  and  the  corresponding  equilibrium  polygon  ACB, 
and,  through  the  points  A,  C,  and  B  draw  a  parabolic 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          55 

curve.  The  bending  moment  diagram,  for  the  highest 
and  lowest  positions  of  the  rod,  will  then  be  the  area  in- 
cluded between  the  curve  and  the  zero  line  AB. 

As  an  example,  take  a  locomotive  with  cylinders  of 
20  inches  diameter  and  26  inches  stroke ;  driving  wheels, 
6  feet  8  inches  diameter ;  side  rods,  7  feet  2  inches 
long  between  centres  and  of  I-section,  2\  x  5f  inches, 
as  shown  in  Fig.  17;  train  speed,  80  miles  per  hour. 

In  substituting  in  the  expressions  for  M  and  ,S,  the 
values  must  be  given  in  the  same  units  of  length  and 
time.  Taking  the  inch  and  the  second  as  these  units, 
<£-=32.2X  12  =  386.4  inches  per  second  per  second; 
V—  80  miles  per  hour  =  1 17.3  feet  per  second  =  1407.6 
inches  per  second  ;  R  —  So/2  =  40  inches  ;  r  =  26/2  = 
13  inches;  v  =  rV/R  —  13  x  1407.6/40  =  457.47  inches 
per  second;  /=  86  inches;  and  the  weight  of  the  steel 
rod  (except  ends)  per  inch  of  length  is  w  =  1.93  pounds. 
The  moment  of  inertia  /  of  the  cross-section  of  the  rod 
about  an  axis  perpendicular  to  the  web  is  29.606 ;  c  = 
2.875  inches;  and  the  section  modulus  T/c=  10.29  and 
its  reciprocal  c/I—  0.097. 

Substituting  in  the  formulas,  we  have :  centrifugal 
force  per  inch  of  length  of  bar  =/=  80.48  pounds  ;  total 
centrifugal  load  =  F—  80.48  x  86  =  6921  pounds ;  bend- 
ing moment  at  mid-length  of  bar  (ordinate  at  C,  Fig. 
17)=  M—  6921  x  -8/-=  74,401  pound-inches;  and  the 
maximum  stress  at  mid-length  =  5  =  74,401  x  0.097  = 
7217  pounds  per  square  inch,  a  result  which  is  low 
owing  to  the  fact  that  the  rod  is  short  and  its  moment 
of  inertia  is  relatively  large.  With  some  rods,  100  inches 


50  GRAPHIC    STATICS 

or  more  in  length,  this  stress  at  this  speed  is  approxi- 
mately 14,000  pounds  per  square  inch. 

It  will  be  understood  that  these  calculations  refer  only 
to  the  stress  produced  by  centrifugal  force.  There  is, 
in  addition,  a  direct  compression  or  tension  in  the  rod, 


4 

0 

0 

0 
Q 

O 

6" 

0 

o 

o 
"o 

0 

"^ 

— 

i- 

_£_ 

0 

0 

4 

JL 

0 

0 

_°£ 

- 

1 

Fit;.  18 

whose  maximum  value,  with  two  driving  wheels  on  each 
side,  is  one-half  of  the  greatest  thrust  on  the  piston. 
The  rod,  when  under  compression,  acts  as  a  column, 
and  its  cross-sectional  area  should  be  sufficiently  great 
to  keep  the  direct  compressive  stress  below  5000  pounds 
per  square  inch.  Finally,  there  may  be  various  indeter- 
minate stresses,  due  to  lack  of  alignment,  and  momen- 
tarily, to  unequal  adhesion  of  the  driving  wheels. 

12.   GIRDER  STAY  WITH  STRESSES  PRODUCED  BY  ITS 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          57 

SUPPORTING  A  CONTINUOUS  BEAM  UNDER  UNIFORM 
LOAD.  Figure  18  represents  a  girder  stay  for  the  top 
of  the  combustion  chamber  of  a  marine  cylindrical 
boiler.  The  stay  is  a  beam  of  25  inches  span  and  is 
formed  of  two  steel  plates,  each  -|  x  j\  inches ;  the 
stays  are  spaced  *j\  inches  apart,  centre  to  centre.  The 
working  pressure  is  160  pounds,  gauge.  The  resultant 
load  on  the  top  of  the  combustion  chamber  is  trans- 
mitted to  each  stay  by  three  bolts,  spaced  6\  inches 
apart,  and  I JJ  inches  diameter  at  the  root  of  the  thread. 
Each  bolt  passes  through  a  yoke-plate  at  the  top  which 
divides  the  load  between  the  two  parts  of  the  stay  and 
also  keeps  them  from  separating.  It  is  required  to  find 
the  maximum  stress  in  the  two  plates  of  the  stay. 

An  approximate  method  which  is  sometimes  used,  is 
to  assume  that  the  stay,  as  a  simple  beam,  carries, 
through  each  of  its  three  bolts,  one-fourth  of  the  total 
load  on  the  strip  of  25  inches  span  and  7|  inches 
width,  the  remaining  fourth  being  divided  equally  be- 
tween the  tube-sheet  and  the  back  plate  of  the  combus- 
tion chamber.  On  this  basis,  we  have,  for  each  stay  : 

Total  load  on  braced  strip  =  25  x  7.5  x  160  —  30,000  Ib. 

Load  on  each  bolt  =  span  load  between  bolts  =  30,000/4  =    7.500  Ib. 

Load  on  tube  sheet  =  load  on  back  plate  —  7500/2  =    3.750  Ib. 

The  problem  then  reduces  to  that  of  a  simple  beam 
AB,  Fig.  1 8,  having  three  concentrated  and  approxi- 
mately equally  spaced  loads,  each  of  7500  pounds,  and 
hence  an  effective  reaction  at  each  support  of  7500  x  f 
=  11,250  pounds.  Constructing  the  force  and  equilib- 


58  GRAPHIC    STATICS 

rium  polygons,  Oab  and  ACB,  the  maximum  bending 
moment  at  C  is  found,  by  measuring  the  ordinate  there, 
to  be  93,750  pound-inches  =  M  =  SI/c  (Art.  10). 

For  a  rectangle  of  base  b  and  depth  d,  7=&/3/12  and 
c  =  d/2.  Hence  I/c  =  bd*/6  and  5  =  6  M/bd\  In  this 
case,  the  two  plates  of  the  girder  are  virtually  combined, 
since  the  yoke-plate  and  the  central  line  of  rivets  pre- 
vent buckling.  Therefore,  b  =  £  x  2  =  1.75  inches, 
d=  7.5  inches,  and 

5  = x  93>75° — =5714  pounds    per    square    inch, 

1.75  x  7-5  x  7-5 

which  is  the  maximum  stress  in  the  girder,  since  the 
ordinate  at  C  gives  the  maximum  bending  moment. 

This  method  is,  as  stated,  approximate.  For  an 
accurate  determination  of  the  stresses,  the  top  of  the 
combustion  chamber  should  be  treated  as  a  series  of 
continuous  beams,  the  end  supports  of  which  are  the 
walls  of  the  chamber,  the  bolts  being  the  intermediate 
supports.  The  reactions  at  these  intermediate  supports 
form  the  loads  on  the  bolts,  and,  when  these  loads  are 
found,  the  girder  may  be  treated,  as  before,  as  a  simple 
beam. 

Referring  to  the  top  plan  of  the  combustion  chamber 
in  Fig.  1 8,  it  will  be  seen  that  there  are  8  equally  spaced 
stays,  each  supporting  a  transverse  section,  7.5x25 
inches,  of  the  top,  which  section  is  virtually  a  continuous 
beam,  supported  from  above.  In  this  section,  there  are 
4  equal  spans,  6\  inches  long,  the  load  on  each  being 
6.25  x  7.5  x  1 60  =  7500  pounds.  In  the  transverse 
direction,  parallel  to  the  stay,  the  top  of  the  chamber 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          59 

will  sag  under  the  load  and  between  the  bolts,  thus 
producing  bending  moments  in  that  direction.  Let 
w  =  span  load  =  7500  pounds.  Then,  according  to  the 
Theorem  of  Three  Moments,  the  reactions  at  the  5  sup- 
ports are : 

Support  12345 

Reaction       iiw/28       32^/28       267^/28       327^/28       nw/28 

It  will  be  seen  that  the  distribution  of  the  load  is  not 
uniform  and  that  the  greater  loads  come  on  the  second 
and  fourth  supports,  i.e.,  on  the  two  outer  bolts,  aa,  of 
the  stay.  Each  of  these  bolts  supports  one-half  of  each 
of  the  adjacent  spans  of  the  transverse  strip,  so  that, 
owing  to  the  bending  stress,  the  area  thus  supported 
may  be  considered  as  having  a  load  of  ii\  =  32  w/2% 
=  8571  pounds.  The  load  on  the  similar  central  area 
and  bolt  is  w2  =  26  w/28  =  6964  pounds,  and  the  total 
load  on  the  5  supports  is  112^/28  =  30,000  pounds,  as 
previously  computed. 

Again,  the  top  of  the  combustion  chamber  will  sag, 
under  the  load  and  between  the  bolts,  in  the  longitudi- 
nal direction,  at  right  angles  to  the  stays.  It  may, 
therefore,  be  considered  as  divided  in  that  direction  into 
3  continuous  beams,  each  with  9  equal  spans  and  10 
supports.  The  reaction  at  support  10  is  the  same  as 
that  at  2,  etc.  Neglecting  the  load  just  computed  for 
transverse  bending,  we  have,  for  the  original  span  load 
w  (  Theorem  of  Three  Moments) : 

Support  i  2  3  4  5 

Reaction,  2097^/530,  6017^/530,  5117^/530,  535^/530,  5297^/530 


60  GRAPHIC    STATICS 

The  greater  loads  are  therefore  those  on  supports  2 
and  9,  which  supports,  for  the  3  beams,  are  the  3  bolts 
in  each  of  the  two  end  girders  of  the  series.  These 
two  stays  are  therefore  those  which  are  under  the  maxi- 
mum stress.  . 

Considering  both  transverse  and  longitudinal  bending, 
the  resultant  loads  on  the  bolts  of  the  two  end  girders 
are : 

On  each  outer  bolt,  a,  load  =  (wx  32/28)  (601/530)  —  1.29^  =  9675^3. 
On  middle  bolt,  b,  load  =  (w  x  26/28)  (601/530)  =  i  .05  w=  7875  Ib. 

That  is,  the  outer  two  of  the  three  continuous  beams 
assumed  for  longitudinal  bending  stresses  are  considered 
as  having  a  span  load  throughout  of  wl  —  8571  pounds, 
and  similarly,  the  middle  beam,  a  uniform  span  load  of 
w^  =  6964  pounds. 

Treating  the  girder  as  a  simple  beam  and  following 
the  approximate  method  in  all  respects,  except  the 
changes  as  above  in  the  bolt  loads,  the  maximum  stress 
5  in  the  girder  is  found  to  be  6686  pounds  per  square 
inch,  or  17  per  cent  more  than  that  given  by  the  ap- 
proximate method.  These  calculations  apply  only  to 
the  two  girder  stays  at  the  ends  of  the  row ;  the  inter- 
mediate girders  are  under  less  stress,  as  is  shown  by  the 
coefficients  for  the  reactions  at  the  corresponding  sup- 
ports. The  values  given  for  these  coefficients  refer 
only  to  beams  of  constant  cross-section,  under  a  uniform 
load,  and  with  all  supports  at  the  same  level.  Strictly 
speaking,  they  are  limited  also  to  beams  with  free  ends 
(Art.  7),  although  the  difference  in  this  respect  is  not 
material  in  this  case. 


STATIONARY  LOADS:    SHEARS  AND  MOMENTS        6 1 

13.   COUNTER  SHAFT:  TWISTING  AND  BENDING    COM- 
BINED.    Figure  19  represents  a  counter  shaft  AB,  sup- 


FIG.  19 


ported  in  bearings  at  A  and  B,  and  carrying  a  26-inch 
driving  pulley  C  and  a  24-inch  driven  pulley  D,  both  of 


62  GRAPHIC    STATICS 

5-inch  face.  The  belt  drive  at  C  is  assumed  to  be  hori- 
zontal; that  at  D,  vertical.  Take  the  vertical  load  at  D 
as  250  pounds,  this  including  the  weights  of  pulley  and 
belting  and  the  pull  of  the  belt  tension.  At  C,  let  the 
vertical  load,  or  weights  of  pulley  and  belting,  be  100 
pounds,  and  the  horizontal  load,  or  belt  tension,  200 
pounds. 

The  shaft  is  bent  in  a  vertical  plane  by  a  force 
7^  =  250  pounds  acting  at  D  and  a  force  P2=  100 
pounds  acting  at  C,  and,  in  a  horizontal  plane,  by  the 
force  PB  =  200  pounds  acting  at  C.  It  is  also  twisted, 
between  C  and  D,  by  a  driving  moment  =  200  x  %f-  = 
2600  pound-inches. 

The  diagram  for  the  vertical  bending  moments  is 
marked  V.  The  moments  shown  by  the  portion  of  the 
diagram  above  the  zero  line  are  positive,  i.e.  the  lower 
side  of  the  shaft  is  in  tension  and  is  bent  in  a  convex 
curve;  the  moments  below  the  line  are  negative,  the 
lower  side  of  the  shaft  being  compressed.  The  reac- 
tions are:  R' v=  150  pounds  and  R" v  =  .200  pounds. 

Diagram  H  gives  the  horizontal  bending  moments. 
The  latter  are  all  positive,  and,  assuming  that  the  pulley 
driving  C  is  in  front  of  it,  the  rear  side  of  the  shaft  is 
in  tension  and  convex.  The  reactions  are:  ^^=33-3 
pounds  and  R"h=2^.^  pounds,  the  two  acting  in  op- 
posite directions. 

The  linear  scale  of  the  original  shaft  diagram  was 
i  foot=  I  inch,  or  12  to  I  ;  the  scale  of  the  force  poly- 
gon was  200  pounds  =  I  inch  ;  and  the  pole  distance  F 
was  i^  inches  =  250  pounds.  Hence  (Art.  12),  the 


STATIONARY   LOADS:     SHEARS   AND    MOMENTS          63 

moment  scale  for  F"and  H  was  12  x  250  =  3000  pound- 
inches  per  inch  of  measured  length  of  any  ordinate. 

The  actual  reactions  at  the  bearings  are  the  resultants 
of  the  vertical  and  horizontal  reactions,  as  above.  Thus, 
as  in  (a\  laying  off  R" v  and  R"h  perpendicular  to  each 
other,  their  resultant  dm  =  R2,  inclined  at  the  angle  0 
to  the  horizontal ;  Rl  can  be  found  similarly,  but  its 
inclination  differs  and  its  line  of  action  lies  on  the  other 
side  of  the  shaft. 

Diagrams  V  and  H  are  combined  to  form  the  result- 
ant bending  moment  diagram  R  by  a  method  (Art.  14) 
similar  to  that  used  for  the  reactions.  Thus,  as  in  (b\ 
let  v  be  the  vertical  bending  moment  at  the  section  D 
of  the  shaft  and  h  the  similar  horizontal  moment ;  their 
resultant  r  is,  on  the  same  moment-scale,  the  correspond- 
ing ordinate  of  diagram  R.  The  latter  diagram  gives 
simply  the  magnitudes  of  the  resultant  ordinates  at  all 
points,  without  regard  to  their  sign ;  it  is  not  a  plane 
surface,  but  is  warped,  since  the  inclination  of  the  ordi- 
nates changes  continually. 

The  diagram  for  the  twisting  moment  between  C  and 
D  is  represented  by  the  rectangle  T,  whose  height  is 
made  equal  to  this  moment,  or  2600  pound-inches,  on 
the  same  moment  scale  as  before.  Diagrams  R  and  T 
are  combined  to  form  diagram  E  for  the  equivalent  bend- 
ing moments  by  the  method  shown  in  (c).  The  equation 
for  the  equivalent  bending  moment  (Art.  14)  is  : 


EMb  = 
in  which  Mb  and  Mt  are  respectively  the  bending  and  twist- 


64  GRAPHIC    STATICS 

ing  moments  at  the  section  considered  and  EMb  is  the 
corresponding  equivalent  bending  moment.  To  apply 
this  expression  graphically  for  any  section,  as  that  at  D, 
lay  off,  as  in  (c\  fg  and  gk  equal  respectively  to  one-half 
the  bending  and  twisting  moments.  Revolve/ to/'  and 
g  to  g1.  Then,  for  the  section  at  D : 

EMb  —  £k  =  ordinate  e. 
This  follows  since,  for  that  section, 
=  I  Mh  an< 


This  method  is  used  for  all  sections  to  the  right  of  D ; 
to  the  left  of  it,  no  twisting  moment  exists  and  the 
original  bending  moment  holds. 

Considering  strength  only,  the  shaft  should  be  de- 
signed for  the  equivalent  bending  moment  shown  by 
the  maximum  ordinate  in  diagram  E.  This  ordinate  is 
that  at  section  D  and  is  equal  to  2516  pound-inches. 
The  resisting  moment  (Art.  10)  of  a  solid  cylinder  to 
bending  strain  is  SI/c,  in  which  5  is  the  maximum  per- 
missible working  stress  and  I/c  =  7rds/$2,  d  being  the 
diameter  in  inches.  Equating  the  equivalent  bending 
and  resisting  moments  and  taking  5  as  16,000  pounds 
per  square  inch  : 

2516=  16,000  x  7n/3/32> 
d=  1.17  inches, 

which  is  less  than  is  customary  in  practice,  since  the 
stiffness  of  the  shaft  and  the  bearing  surface  at  the 
journals  must  be  considered,  as  well  as  its  strength  to 
resist  bending  and  shearing.  The  small  diameter  of 


STATIONARY    LOADS:     SHEARS    AND    MOMENTS          65 

the  shaft  in  this  case  is  produced  mainly  by  the  rela- 
tively low  value  of  the  bending  moment,  and  this  low 
value  is  due  to  the  fact  that  the  shaft  is  short  and  the 
weight  of  pulley  C  acts  to  balance  that  of  pulley  D. 


—  - 14  - 


FIG.  20 


14.  CENTRE  CRANK  SHAFT:  TWISTING  AND  BENDING 
COMBINED.  Figure  20  represents  the  centre  crank 
shaft  of  a  vertical,  high  pressure  engine,  the  shaft  being 
supported  in  bearings  at  A  and  B  and  delivering  its 


66  GRAPHIC    STATICS 

power  as  a  twisting  moment  at  the  left-hand  end.  For 
simplicity,  neglect  the  weight  and  friction  of  the  shaft 
and  reciprocating  parts.  It  is  required  to  find  the 
equivalent  bending  moments  on  the  shaft  and  cranks 
when  the  engine  is  developing  its  maximum  power,  i.e., 
when,  as  in  {A),  the  connecting  rod  and  cranks  are  at 
right  angles  on  the  downward  stroke. 

The  data  are:  cylinder  diameter,  15  inches;  stroke, 
1 8  inches;  length  of  connecting  rod,  29  inches;  maxi- 
mum unbalanced  pressure  on  piston,  100  pounds  per 
square  inch;  force  on  piston  at  position  shown  in 
(A}  =  Q  —  TT  x  15x15  x  J  x  100=17,671  pounds;  cor- 
responding thrust  on  connecting  rod  =  P  =  Q/cos  0  = 
17,671  x  1.04=  18,400  pounds,  about. 

In  the  position  shown  in  (A),  the  thrust  P  is  resisted 
by  an  equal,  opposite,  and  parallel  reaction  Rt  whose 
line  of  action  passes  through  the  centre  of  the  shaft. 
To  simplify  the  diagrams  to  be  drawn,  the  forces  P  and 
R,  considered  as  bending  forces  only,  are  assumed  to  be 
revolved  through  90  degrees,  as  shown  by  dotted  lines, 
and  hence  to  act  in  the  plane  of  the  crank  axes — an 
assumption  which  evidently  does  not  alter  the  magni- 
tude of  the  bending  action  of  these  forces  on  the  shaft 
and  crank  pin.  With  regard  to  the  bending  of  the 
crank  and  the  twisting  of  the  latter,  of  the  crank  pin, 
and  of  a  portion  of  the  shaft,  the  two  forces  are  con- 
sidered as  acting  in  their  original  planes. 

Let  abcdef  represent  the  neutral  axes  of  the  shaft, 
cranks,  and  crank  pin,  i.e.  a  line  coinciding  with  the 
axes  of  the  shaft  and  pin  and  passing  through  the 


STATIONARY   LOADS:    SHEARS   AND   MOMENTS          67 

centres  of  gravity  of  all  transverse  sections  of  the 
cranks. 

The  shaft  is  subjected  to  bending  from  a  to  b  and 
from  e  to  /,  and  to  twisting  from  b  to  a  and  onward  to 
the  left;  the  crank  pin  is  under  both  bending  and 
twisting  strains  throughout  its  entire  length  ;  and  both 
cranks  are  twisted  on  their  neutral  axes  and  are  also 
bent  as  cantilevers  (Art.  7). 

On  the  load-line  gh,  lay  off  P  and  draw  the  force 
polygon  Ogh  and  the  bending  moment  diagram  aDf. 
The  ray  Og' ,  parallel  to  the  closing  line  af,  determines 
the  magnitude  of  the  reactions,  which  are  :  Rl  =  11,040 
pounds  and  R2  =  7360  pounds.  With  a  linear  scale  of 
8  to  i,  a  force  scale  of  8000  pounds  per  inch,  and  a 
pole  distance  of  i^  inches,  the  moment  scale  (Art.  12) 
is  96,000  pound-inches  per  inch. 

Shaft.  The  right-hand  section  of  the  shaft  acts  simply 
as  a  support  and  is  therefore  under  bending  strain  only, 
the  moments  being  given  by  the  portion  fek  of  the 
diagram. 

The  similar  bending  moment  diagram  for  the  left- 
hand  section  is  abm.  This  section  is  also  subjected  to 
twisting  by  the  driving  moment,  P  x  r—  18,400  x  9  = 
165,600  pound-inches.  Using  the  same  moment  scale 
as  above,  the  twisting  moment  diagram  is  bnoa.  Com- 
bining the  two  diagrams  by  the  method  employed  in 
Example  13,  the  equivalent  bending  moment  diagram 
is  bn'd a. 

Crank  Pin.  The  crank  pin  and  the  parts  of  the 
cranks  lying  between  the  two  neutral  axes  of  the  latter 


68  GRAPHIC    STATICS 

are  under  bending  strain,  the  moments  of  which  are 
given  by  the  portion  bmDke  of  the  bending  moment 
diagram. 

The  pin  is  also  twisted  by  a  moment, 

R2  x  r  =  7360  x  9  =  66,240  pound-inches. 

This  follows  since  if,  in  the  original  plane  of  the  right 
reaction  R%,  there  be  applied  at  e  two  equal  and 
opposite  forces  R2f  and  R2",  each  equal  to  R%,  equilib- 
rium will  still  exist,  but  the  forces  R2  and  R2  will 
form  a  couple  of  arm  ef  tending  to  twist  the  crank  on 
its  neutral  axis,  and  the  force  R2"  =  R2  will  act,  with 
the  leverage  ed  =  r,  to  twist  the  crank  pin.  These 
forces  are  shown  in  the  diagram  as  if  they  were  in  the 
plane  of  the  paper ;  they  really  act  in  the  plane  of  R  as 
in  (A). 

The  twisting  moment  diagram  is  cpqd.  Combining 
this  with  diagram  bmDkey  we  have  the  equivalent  bend- 
ing moment  diagram  cp'd'q'd.  Since  the  pin  is  cylin 
drical,  the  maximum  ordinate  of  the  latter  diagram,  that 
at  the  centre,  is  the  only  one  used  in  designing  the  pin. 
The  equivalent  bending  moment  corresponding  with 
this  ordinate  is  159,360  pound-inches. 

Crank  C2.  As  has  been  shown,  the  right-hand  crank 
is  subjected  to  twisting  on  its  neutral  axis  ed  and  to 
bending  as  a  cantilever. 

The  bending  is  due  to  the  fact  that  the  crank  pin  is 
twisted,  through  the  medium  of  the  crank,  by  the  force 
R2  acting  at  e.  The  bending  moment  at  the  outer  end 
d  of  the  crank  is  therefore  R2  x  r=  twisting  moment 


STATIONARY    LOADS:     SHEARS    AND    MOMENTS          69 

on  crank  =  66,240  pound-inches;  at  the  inner  end  e, 
the  bending  moment  is  zero.  Revolving  dq  to  dr,  the 
bending  moment  diagram  for  the  crank  is  edr,  the  ordi- 
nates  being  horizontal. 

As  explained  previously,  the  crank  is  twisted  on  its 
neutral  axis  by  a  couple  of  force  Rz  and  of  arm  ef' 
The  twisting  moment  is  therefore  R%  X  ef.  Considering 
the  right  reaction  and  not  the  left  as  previously,  the 
bending  moment  ek  at  the  section  e  is  equal  to  the 
product  of  the  right  reaction  by  the  distance  from  e  to 
the  right  support,  or  R2  x  ef,  which  is  also  the  twisting 
moment  as  above.  Revolving  ek  to  ek1 ,  the  twisting 
moment  diagram  is  the  rectangle  ek'r'd.  Combining 
the  two  diagrams,  we  have  the  equivalent  bending  mo- 
ment diagram  ek"r"d.  The  maximum  ordinate  dr"  of 
this  diagram,  measured  by  the  moment  scale,  gives  an 
equivalent  bending  moment  of  100,800  pound-inches. 

Crank  Cv  The  left-hand  crank  C^  is  subjected  to 
similar  bending  and  twisting  actions. 

With  regard  to  bending  strain,  the  crank  is  a  cantilever, 
fixed  at  b  and  carrying  the  load  P  at  c.  The  bending 
moment  at  the  inner  end  b  is  therefore  P  x  r—  165,600 
pound-inches,  which  is  the  twisting  moment  bn  on  the 
left-hand  section  of  the  shaft;  the  moment  at  the  inner 
end  c  is  zero.  Revolving  bn  to  bn,  the  bending  mo- 
ment diagram  is  cbu,  the  ordinates  being  horizontal. 

By  the  same  reasoning  as  that  followed  for  the  right- 
hand  crank,  it  can  be  shown  that  crank  C^  is  twisted  on 
its  neutral  axis  be  by  the  left  reaction  R^  acting  with 
the  leverage  ab.  The  twisting  moment  is  therefore 


70  GRAPHIC    STATICS 

R1  x  ab,  which  moment  is  equal  to  bm,  the  bending 
moment  on  the  shaft  at  the  section  b.  Revolving  bm 
to  bm' ,  the  twisting  moment  diagram  is  the  rectangle 
bm'c'c. 

The  combination  of  the  two  diagrams  gives  the  equiv- 
alent bending  moment  diagram  bm"c"c.  The  maxi- 
mum ordinate  bm"  of  this  diagram  corresponds  with  a 
bending  moment  of  180,000  pound-inches. 

Diameter  of  Crank  Pin.  The  maximum  equivalent 
bending  moment  on  the  crank  pin  has  been  found  to 
be  159,360  pound-inches,  which  is  the  value  of  M  in 
the  formula,  M—  SI/c.  If  the  pin  be  a  solid  cylinder 
of  diameter  dlt  the  rectangular  moment  of  inertia 
/=  Trd-f/64.  and  c  —  d-^/2.  Taking  a  suitable  value  for 
the  maximum  stress  S,  the  value  of  dl  can  be  deter- 
mined as  in  Example  13. 

If  the  crank  pin  be  of  steel  and  made  hollow  for 
lightness,  the  method  is  the  same,  except  that  the  value 
of  /  is  changed,  being  now  the  difference  between  the 
moments  of  inertia  of  the  inner  and  outer  circular  sec- 
tions. For  example,  if  the  external  diameter  be  d^  and 
the  internal  diameter  d^,  the  value  of  /is  Tr(d^  —  d^}/6^. 
and  the  area  of  the  cross-section  of  the  pin  is 
A  =  Tr(df  —  4z2)/4'  Dividing  : 

I  =  di  +  42 

A  16      ' 


Making   d%  —  d-^/2  =  c   and  5  =  10,000  pounds  per 
square  inch, 


STATIONARY   LOADS:     SHEARS   AND   MOMENTS          71 


1 0,000 
dl  =  5.6  inches,  about. 

The  maximum  shearing  stress  S,  on  the  pin  can  be  as- 
certained from  the  formula  R%  x  r=  S8J/c  (Example 
13),  in  which  the  twisting  moment,  R2  x  r=  66,240  pound- 
inches  and  the  polar  moment  of  inertia,  J=A(d-f  +  d£)/%. 
With  dl  =  6  inches  and  d^  =  c  =  3  inches,  J  =  1 19  and 
S8  is  1670  pounds  per  square  inch,  a  stress  which  is 
immaterial.  The  diameter  of  the  shaft  can  be  found 
similarly;  its  twisting  moment  is  greater,  and  its  maxi- 
mum equivalent  bending  moment  less,  than  those  of  the 
crank  pin. 

With  steel  having  an  elastic  limit  of  50,000  pounds 
per  square  inch,  the  factor  of  safety  for  the  crank  pin 
is  50,000/10,000=5.  The  diameters  as  computed,  6 
and  3  inches,  external  and  internal,  would  serve  for  both 
a  hollow  shaft  and  a  hollow  pin,  if  the  shaft,  cranks, 
and  pin  were  forged  in  one  piece.  If  the  crank  shaft 
be  built  up,  keys  would  be  required  for  both  pin  and 
shaft,  and  the  external  diameters  should  be  increased 
to  allow  for  the  keyways. 

Dimensions  of  Cranks.  The  cranks  should  be  dupli- 
cates and  should  be  designed  for  the  maximum  moments, 
which  are  given  by  bm"  and  dr",  at  the  inner  and 
outer  ends,  respectively.  As  a  rule,  the  cross-section  is 
rectangular  and  the  thickness  t  of  the  web  is  constant. 
For  uniform  strength  throughout,  the  transverse  width 


72  GRAPHIC    STATICS 

w  should  vary  with  the  maximum  moments,  as  given  by 
either  of  the  two  equivalent  bending  moment  diagrams 
—  a  refinement  which  is  possible  only  with  a  casting. 

The  hubs  on  crank  pin  and  shaft  should  be  long 
enough  for  adequate  keying  and  to  prevent  the  crank 
from  jarring  loose.  In  built  up  shafts,  this  factor  usu- 
ally determines  the  thickness  /,  which  is  made  uniform 
throughout  and  equal  to  the  length  of  the  hub,  while 
the  width  w  is  also  uniform  and  equal  to  the  maximum 
external  diameter  required  for  the  larger  of  the  two 
hubs.  In  any  event,  the  maximum  stress  at  any  section 
of  a  crank  of  rectangular  cross-section  can  be  deter- 
mined by  proper  substitution  in  the  general  formula, 
M—  SI/c.  M  is  here  the  maximum  bending  moment 
at  that  section,  as  given  by  the  equivalent  bending 
moment  diagram ;  6"  is  the  corresponding  maximum 
stress  which  occurs  at  either  of  the  outer  ends  of  the 
width  w\  /=/2£^/i2;  and  c=w/2.  The  values  for 
/  and  w  are,  in  each  case,  taken  from  the  dimensions- of 
the  section  under  consideration. 


CHAPTER    IV 

LIVE  LOADS:    SHEARS  AND  MOMENTS 

The  shear  and  bending  moment  diagrams  which  have 
been  considered  thus  far  represent  these  functions  for 
all  sections  of  a  beam  carrying  stationary  loads  only. 
This  limitation  applies  also  when  these  diagrams  have 
been  constructed  for  the  moving  parts  of  machines, 
since  the  members  in  motion  have  been  assumed  to  be 
at  rest  for  an  instant  and  in  momentary  equilibrium 
under  the  action  of  the  driving,  resisting,  and  support- 
ing forces  ;  and,  further,  the  shears  and  moments  which 
have  been  thus  determined,  refer  only  to  the  given 
position  of  the  moving  parts. 

While  the  general  effects  of  a  live  or  travelling  load 
—  as  a  train  passing  over  a  bridge  —  are,  at  any  instant, 
precisely  the  same  as  those  of  an  equivalent  stationary 
or  dead  load  in  the  same  position,  there  is,  so  far  as  the 
maximum  stresses  in  the  members  are  concerned,  an 
important  difference  between  the  two  cases,  in  that, 
with  every  change  in  the  location  of  the  moving  load 
there  are  corresponding  changes  in  the  shears  and  mo- 
ments which  that  load  produces.  Hence,  in  each  mem- 
ber of  a  structure  traversed  by  a  moving  load,  there  is  a 
continual  variation  of  stress,  the  maximum  range  of 

73 


74 


GRAPHIC    STATICS 


which  must  be  determined  in  order  that  the  member 
shall  be  proportioned  for  adequate  strength  under  all 
conditions  of  service.  These  variations  in  the  shears, 
moments,  and  their  corresponding  stresses  will  now  be 
examined. 

15.  Variation  of  Live  Load  Shear  at  Sections  to  the 
Left  of  the  Load,  (a)  Concentrated  Loads.  Let  Fig. 
21  represent  a  simple  beam  of  span  s,  traversed  by  a 


FIG.  21 

single  concentrated  load  W.  The  vertical  shear  V 
(Art.  8)  is  the  resultant  or  algebraic  sum  of  all  forces 
to  the  left  of  the  section  considered.  With  a  single 
moving  load,  as  in  this  case,  Fis  therefore  equal  to  the 
left  reaction  /?r  Assume  that  the  load  is  moving  from 
right  to  left,  and  let  x  be  the  distance  from  the  right 
support.  Taking  moments  about  the  right  support  : 

W  y.  x  =  R  x  s 


which  is  the  vertical  shear  for  all  sections  between  the 
points  A  and  B,  and  to  the  left  of  the  load. 


LIVE  LOADS:   SHEARS  AND  MOMENTS  75 

This  expression  is  the  equation  of  a  straight  line  db 
inclined  to  the  horizontal  by  the  angle  whose  tangent 
is  W/s.  When  x  =  s,  Rl=V=W;  when  ;r  =  o, 
R±  =  V—  o.  Therefore,  the  horizontal  line  ac  is  the 
zero  line,  and,  at  any  position  of  the  load,  the  ordinate^ 
just  below  the  load  —  or,  strictly  speaking,  immediately 
to  the  left  of  its  centre  of  gravity  —  and  included  in 
the  diagram  abc  will  give  Ffor  that  section,  and  similarly 
for  all  sections  to  the  left  of  W.  It  is  evident  that  this  is 
true,  whether  the  load  be  moving  to  the  right  or  left, 
since,  in  this  case,  we  have  but  one  concentrated  load 
to  consider.  The  diagram  abc,  therefore,  represents  the 
variation  of  the  vertical  shear  at  all  sections  to  the  left 
of  the  load. 

For  sections  to  the  right  of  the  load,  V—  R1  —  W,  an 
expression  which  is  the  same  as  the  preceding  equation, 
except  that  there  is  a  constant  deduction  W.  It  is, 
therefore,  the  equation  of  the  straight  line  cd,  which 
is  parallel  to  bat  and  starts  at  the  point  c  of  the  zero 
line.  Hence,  the  diagram  cad  represents  the  variation 
of  the  shear,  now  negative,  for  the  sections  to  the  right 
of  the  load. 

(b)  Uniform  Load.  Now,  consider  a  moving  uniform 
load,  long  enough  to  cover  the  span  when  located  be- 
tween the  two  supports  of  the  beam.  In  practice,  the 
cars  of  a  railroad  train,  although  not  the  locomotive  and 
its  tender,  are  considered  as  such  a  load  in  passing  over 
a  bridge.  This  case  differs  from  the  one  just  discussed, 
in  that  the  total  load  on  the  beam  varies  with  the  length 
of  the  segment  of  the  span  covered  by  the  load. 


76 


GRAPHIC    STATICS 


Let  Fig.  22  represent  a  simple  beam  of  span  s,  carry- 

ing a  partial  uniform  load   Wt  composed  of  unit  loads 

spaced  at  a  unit  distance  apart.     If  x  be  the  number  of 

unit  distances  covered  by  the  load  at  a  given  time  and 

ft__  _  5  _  w  De  the  weight 

of  the  unit  load, 
then  the  total 
»  §  load  at  that  time 
will.be  wx  •  =  W, 
which  is  there- 
fore variable. 
This  total  load, 
or  resultant  of 
all  of  the  unit 
loads,  will  be 
taken  as  concen- 
trated at  the 
centre  of  gravity  of  the  uniform  load.  It  is  required  to 
determine  the  variation  in  the  vertical  shear  for  all  sec- 
tions of  the  beam  to  the  left  of  the  head  of  this  moving 
uniform  load. 

Assume  the  load  to  be  travelling  from  right  to  left. 
For  all  sections  to  the  left  of  the  head  H  of  the  load, 
V=  R^  Taking  moments  about  the  right  support  : 

^  x  s  =  W  x  x/2  =  w*2/2, 


FIG.  22 


which  is  the  equation  of  a  parabola.     If  x  =  o,  V=  o  ; 
if  x  —  s,  V=  ws/2.     Plotting  ab  as  a  parabolic  curve  in 


LIVE  LOADS:   SHEARS  AND  MOMENTS  77 

accordance  with  the  equation  for  V,  the  diagram  abc 
represents  the  variation  of  V,  for  all  sections  to  the  left 
of  Ht  while  the  load  of  final  length  s  moves  from  right  to 
left.  Wherever  H  may  be  located,  the  ordinate  y  im- 
mediately below  it  will  give  the  value  of  V  from  H  to 
the  left  support,  for  that  position  of  the  load. 

Again,  assume  the  load  to  be  moving  from  left  to 
right  and  measure  x  from  the  left  support.  Taking 
moments  about  the  right  support: 

R^  X  S  =    W(S  —  X/2)  —  WX(S  —  X/2\ 

Rl  =  wx  —  zvx2/2s, 
V=  Rl  —  W—  wx  —  wx^/2  s  —  wx  —  —  wx^/2  s, 

which  is  the  same  equation  as  before,  except  that  V  is 
now  negative.  If  x  —  o,  V—Q\  if  x=s,  V=  —ws/2. 
Hence,  the  diagram  cda  represents  the  variation  of  V 
for  all  sections  between  H  and  the  left  support,  the 
ordinate  y  under  H  giving  the  required  magnitude  of 
F,  which  is,  in  all  cases,  negative. 

16.  Influence  Diagrams :  Influence  Lines.  The  line 
ab  and  the  diagram  abc,  Fig.  21,  are  termed  an  influence 
line  and  an  influence  diagram,  respectively.  In  this 
case,  since  V—R^,  they  represent  the  variation  of  both 
the  vertical  shear  and  the  left  reaction.  Influence 
diagrams  are  usually  constructed  on  the  basis  of  the 
unit  load  —  a  pound,  kilogram,  or  ton.  The  ordinates 
of  the  diagram  then  refer  to  the  effect  of  that  unit  load 
only,  and,  to  obtain  the  actual  reaction,  shear,  etc.,  for 
the  given  load,  the  length  of  each  ordinate  is  multiplied 
by  the  number  of  pounds,  tons,  etc.,  in  that  load. 


GRAPHIC   STATICS 


17.  Variation  of  Live  Load  Shear  at  any  Given 
Section  of  a  Beam,  (a)  Concentrated  Load.  Let  Fig. 
23  represent  a  single  concentrated  load  W>  moving 


FIG.  23 

from  right  to  left  over  a  simple  beam  of  span  s.  It 
is  required  to  find  the  variation  in  the  shear  at  the 
section  //,  distant  m  units  from  the  left  support,  while 
the  load  traverses  the  beam. 

Assume  the  load  to  be  to  the  right  of  H  and  distant  x 
units  from  the  right  support.  Taking  moments  about 
that  support : 

W  x  x  —  R   x  s, 


which  is  the  equation  of  the  straight  line  ab,  inclined  to 
the  horizontal  by  the  angle  whose  tangent  is  W/s.  If 
x  =  o,  V=o\  iix—s,  V=  W.  Hence,  ac  is  the  zero 
line,  and  the  diagram  abc  gives  the  magnitude  of  V 
for  all  sections  between  W  and  the  left  support ;  and, 
for  the  section  at  H,  the  ordinate  y,  immediately  under 


LIVE  LOADS:   SHEARS  AND  MOMENTS  79 

the  load  and  included  within  the  partial  diagram  ade, 
represents  V,  while  W  is  to  the  right  of  H. 

When  W  is  between  H  and  the  left  support,  Rl  is 
still  equal  to  Wx/st  but  V=  R1  —  W  and  is  negative,  so 
that  the  line  cf  is  parallel  to  ba  and  begins  at  the  point 
c,  where  x—s  and  V=o.  Hence,  the  ordinate  immedi- 
ately below  the  load  and  included  within  the  partial 
diagram  gee 
gives  the  shear  "* 


i  """  ~  ~      ~  """ 

between     the       1 


?  1  H  A 

P,  f*--  n  —  >j 

—  m§  ----  i 


load     and     the 

left    support, 

which    shear   is  ^--j 

always  negative,  w  j 

The      complete 

diagram,   repre- 

senting the  vari- 

ation of  Fat  the  FIG.  24 

section  H  for  all 

positions  of  the  load,    is   adegce.     This  diagram  serves 

whether  the  load  moves  toward  the  right  or  the  left. 

(b)  Uniform  Load.  Let  Fig.  24  represent  a  simple 
beam  of  span  s,  traversed  from  right  to  left  by  a  uni- 
form load  of  w  pounds  per  unit  of  length  of  the  span, 
the  load  being  long  enough  to  cover  the  span.  It  is 
required  to  find  the  variation  in  shear  at  the  section  H, 
distant  m  units  from  the  left  support,  while  the  head  of 
the  load  crosses  the  span. 

Draw  the  zero  line  ab  =  s  ;  from  b,  erect  be,  and,  from 
ay  let  fall  ad,  both  equal  to  w  ;  draw  ac  and  bd.  Let 


8o  GRAPHIC    STATICS 

x=  length  of  the  load,  as  measured  from  the  right  support. 
Then,  as  shown  previously,  while  the  head  of  the  load  is 
to  the  right  of  section  H,  the  shear  at  that  section  is 
positive  and  is  : 


V=  Rl  =  wx*/2  s. 

With  the  head  of  the  load  at  H,  this  expression  is  the 
area  of  the  triangle  aef,  forming  the  part  of  the  diagram 

below  the  load,  since  : 

ef  :  x  :  :  w  :  s, 

ef—  wx/2, 

area  aef=  ef  x  xJ2  =  zux^/2  s  =  V. 
When  the  lead  moves  to  the  section  If',  distant  n 
units  from  H,  the  shear  at  section  H  is  the  resultant  of 
the  positive  shear  produced  by  the  partial  load  of  length 
x  —  n  (x  being  now  the  distance  from  the  right  support 
to  the  point  H')  and  the  negative  shear  due  to  the  load 
of  length  n,  both  in  the  positions  shown.  This  resultant 
shear  is  the  difference  between  the  areas  of  the  positive 
and  negative  sections  of  the  diagram  below  the  load, 
or  area  aef  minus  fghk.  As  before,  Rl  =  wx^/2  s,  but 
the  shear  at  section  His  now  equal  to  R1  minus  the  load 
nw  to  the  left  of  H,  or 

F=  Rl  —  nw  =  wx*/2  s  —  nw. 

With  the  head  of  the  load  at  If',  wx?/2  s  is  the  area 
of  the  triangle  akp,  while  nw  is  the  area  of  the  parallelo- 
gram eghp.  Hence 

area  akp  —  area  nw  =  area  aef—  zxv&fghk. 

18.  Maximum  Live  Load  Shear.  From  Fig.  24,  it  will 
be  seen  that,  when  the  given  section,  as  that  at  If',  divides 


LIVE  LOADS:   SHEARS  AND  MOMENTS  81 

the  span  into  two  unequal  segments,  the  maximum  numer- 
ical value  of  the  live  load  shear  due  to  a  uniform  load  will 
occur  when  the  load  covers  fully  and  only  the  greater 
segment  of  the  span,  and  that  this  value  will  grow  with 
increase  in  the  length  of  the  segment,  as  compared  with 
that  of  the  span.  If  this  greater  and  loaded  segment  lies 
to  the  right  of  the  section  considered,  the  maximum  live 
load  shear  will  be  positive ;  if  to  the  left,  negative. 

With  a  series  of  concentrated  loads  differing  in  mag- 
nitude and  unequally  spaced,  the  principle  as  to  the 
loading  of  the  greater  segment  of  the  span  to  obtain  the 
maximum  live  load  shear  still  holds,  except  that  the  maxi- 
mum shear  may  occur  when  one  or  more  of  the  loads 
have  passed  beyond  the  section  considered  and  are  sup- 
ported by  the  smaller  segment  of  the  beam.  This  fol- 
lows from  the  fact  that  the  value  of  the  shear  depends 
primarily  on  that  of  the  left  reaction,  and  that  the  value 
of  the  latter  grows  with  every  added  load  borne  by  the 
beam.  Hence,  with  some  of  the  loads  on  the  lesser 
segment,  the  shear,  or  algebraic  sum  of  the  increased 
reaction  and  these  loads,  may  be  greater  than  the  reac- 
tion only  with  the  smaller  number  of  loads  which  the 
length  of  the  larger  segment  will  admit.  The  final 
shear  at  any  section  is,  in  any  case,  the  algebraic  sum 
of  those  due  to  the  dead  and  live  loads. 

19.  Counterbracing.  The  upper  diagram,  Fig.  25,  is 
an  outline  of  a  bridge  truss*  of  the  Pratt  type,  in  which 

*  Bridge  trusses  are  built  either  as  deck  or  through  spans.  In  the  former, 
the  roadway  rests  on  the  upper  chord;  in  the  latter,  on  the  lower  chord. 


82 


GRAPHIC   STATICS 


V(-) 


(a) 


FIG.  25 


the  diagonal  members  are  built  to  withstand  tension 
only.  As  shown  in  Fig.  24,  a  uniform  load,  like  a  train 
of  cars  crossing  a  bridge  truss,  produces  positive  shear 
in  moving  from  the  right  abutment  to  the  left  and  nega- 
tive shear  in  passing  in 
the  other  direction.  The 
dead  load  shear  is  pos- 
itive from  the  left  abut- 
ment to  the  middle  of 
the  truss  and  negative 
from  the  middle  to  the 
right  abutment.  With 
the  train  coming  from 
either  direction  and  pass- 
ing over  a  5-panel  truss,  like  that  shown  in  Fig.  25, 
the  resultant  of  the  shears  due  to  dead  and  live  loads 
is  always  positive  in  the  first  and  second  panels 
(counting  from  the  left),  and  always  negative  in  the 
fourth  and  fifth  panels;  but,  in  the  middle  panel,  this 
resultant  shear  is  positive  when  the  train  comes  from 
the  right  and  negative  when  it  enters  from  the 
left. 

The  vertical  shear  which  exists  in  a  panel  is  a  force 
which  acts  upward  if  the  shear  be  positive,  and  down- 
ward if  it  be  negative.  In  the  Pratt  truss,  under  any 
system  of  loading,  the  vertical  members  are  always  in 
compression,  and  the  diagonals  are  built  to  take  tension 
only.  Hence,  when  the  shear  in  the  middle  panel 
changes  from  positive  to  negative,  or  the  reverse,  there 
must  be  a  corresponding  shifting,  from  one  diagonal  to 


LIVE  LOADS:   SHEARS  AND  MOMENTS  83 

the  other,  of  the  tensile  stress  whose  vertical  component 
is  to  resist  this  shear,  i.e.  the  main  tie  (diagonal)  ceases 
to  act  and  the  counter-tie  takes  the  load,  or  vice  versa, 
the  two  ties  being  oppositely  inclined. 

Thus,  in  (a),  Fig.  25,  the  shear  in  the  middle  panel  is 
positive ;  the  stress  in  the  vertical  5-4  is  compressive, 
and  therefore  acts  toward  panel-point  4 ;  and  the  stress 
in  the  main  diagonal  4-7  is  tensile,  and  acts  from  point 
4.  The  panel  is  subjected  to  the  upward  force  V,  and 
the  only  vertical  forces  at  point  4  are  the  compressive 
stress  3-4  and  the  vertical  component  of  the  tensile 
stress  4-7.  For  equilibrium,  these  vertical  forces  must 
be  equal  and  opposite.  Hence,  Fis  equal  in  magnitude 
to  stress  4-5,  and  also  to  the  vertical  component  of  the 
stress  4-7,  and  is  the  same  in  direction  as  stress  3-4 
with  regard  to  panel-point  4.  When,  as  in  (b\  the 
shear  in  the  panel  becomes  negative,  V  is  equal  and 
opposed  to  the  vertical  component  of  the  stress  in  the 
counter-tie  5-6 1  and  is  equal  and  like  the  stress  4-3  in 
the  vertical,  both  with  regard  to  panel-point  5. 

When  the  main  diagonal  is  acting,  the  counter-tie 
buckles  and  is  not  under  stress ;  owing  to  their  opposite 
inclination,  both  ties  cannot  be  strained  simultaneously. 
This  is  the  fundamental  principle  of  counterbracing, 
which  is  applied  whenever  a  member  is  subject  to  stress 
reversal  and  is  fitted  to  take  tension  only.  In  the  War- 
ren truss,  the  members  are  built  for  both  stresses,  and 
counterbraces  are  not  required.  In  roof  trusses  the 
wind  loads  may  cause  a  reversal  of  stress  in  some  mem- 
bers, thus  necessitating  counterbracing. 


04  GRAPHIC    STATICS 

20.  Variation  of  Bending  Moment  at  any  Given  Sec- 
tion of  a  Beam,  (a)  Concentrated  Loads.  Let  Fig.  26 
represent  a  simple  beam  of  span  s,  traversed  from  right 
to  left  by  a  single  concentrated  load  W.  It  is  required  to 
find  the  variation  in  the  bending  moment  at  the  section 


FIG.  26 


A,  distant  s1  from  the  left  support  and  sz  from  that  at 
the  right,  while  the  load  crosses  the  beam. 

The  bending  moment  (Art.  9)  at  any  section  is  the 
moment  of  the  left  reaction  about  any  point  in  that  sec- 
tion, minus  the  similar  moments  of  the  loads,  if  any,  to 
the  left  of  the  section.  First,  let  the  load  W  be 
between  the  section  A  and  the  right  support,  at  the  dis- 
tance^ from  the  latter.  The  left  reaction  is  R1  = 
and  the  bending  moment  at  section  A  is  : 


J/o 


Wsl  x 


LIVE  LOADS:   SHEARS  AND  MOMENTS  85 

which  is  the  equation  of  the  straight  line  ab,  inclined  to 
the  horizontal  by  the  angle  whose  tangent  is  s-^/s,  the 
weight  W  being  taken,  for  convenience,  as  a  unit  load. 
Now,  assume  the  load  Wto  be  between  the  section  A 
and  the  left  support,  at  the  distance  xl  from  the  latter. 
Then,  Rl  =  W(s—^l)/s,  and  the  bending  moment  at 
section  A  is  : 


Taking  ^as  the  unit  load,  this  expression  is  the  equa- 
tion of  the  straight  line  cd,  inclined  to  the  horizontal  by 
the  angle  whose  tangent  is  s%/s. 

The  triangle  aec  is  the  influence  diagram  for  the 
bending  moments  at  section  A,  when  W  is  equal  to  the 
unitload,  and  the  ordinate  y1  or  j/2  immediately  below 
the  load  gives  the  bending  moment  at  A  for  that  posi- 
tion of  the  load.  This  follows,  since  : 


and 


j2  =  s1  x  .r2/s  =  Mr 

The  diagram  has  been  constructed  for  the  unit  load. 
To  obtain  the  bending  moment  for  a  load  IV,  greater 
than  unity,  each  ordinate  should  be  multiplied  by  the 
number  of  units  in  W. 

(b)  Uniform  Load.  The  influence  diagram  in  Fig^ 
26  is  drawn  for  the  unit  load.  With  a  total  uniform 
load  W  covering  a  segment  or  all  of  the  span,  there  is 
a  series  of  unit  loads,  each  equal  to  wt  spaced  at  unit 


86  GRAPHIC    STATICS 

distance  apart,  over  the  segment  covered.  The  bending 
moment  due  to  each  unit  load  is  given  by  the  ordinate 
immediately  below  that  load.  Hence,  the  total  bending 
moment  produced  by  the  total  load  W  is  the  sum  of 
these  unit  moments,  or  the  area  of  the  portion  of  the 
influence  diagram  below  the  total  load  W. 

Thus,  in  Fig.  26,  let  the  uniform  load  be  moving 
from  right  to  left  and  let  the  head  of  the  load  be  at  the 
distance  xz  from  the  right  support.  Taking  moments 
about  that  support  : 

R^  x  s2  =  wx2  x  ^2/2  ; 
and  the  left  reaction  is  : 

R^  =  wx^/2  s. 

The  bending  moment  at  the  section  A  due  to  the  total 
load  W=  ivx,  is  : 

R1  xs1  =  ws^xf/2  s  =  M. 
But: 


and  the  area  of  the  triangle  below  the  load  is  : 

j2  x  ^2/2  =  V22/2  s=Mt 
since  wt  as  the  unit  load,  is  equal  to  unity. 

21.   Maximum  Bending  Moments  due  to  Live  Loads. 

The  maximum  bending  moment  which  is  possible  at 
any  given  section  of  a  beam  under  any  system  of  load- 
ing, depends  on  two  conditions  :  first,  the  value  of  the 
moment  rises  with  increase  in  the  magnitude  of  the  left 
and,  second,  this  value  is  lessened  by  increase 


LIVE  LOADS:   SHEARS  AND  MOMENTS  87 

in  the  moments  of  any  loads  to  the  left  of  the  section. 
Hence : 

(a)  With  uniform    live  loads,  the  maximum  moment 
at  any  section  of  a  beam  occurs  when  the  load  covers 
the  entire  span. 

(b)  With  concentrated  live  loads,  unequal  in  magni- 
tude and  unequally  spaced,  the  maximum  moment   at 
any  section  occurs  when  the  span  carries  the   greatest 
possible  load,  consistent  with  having  one  of  the  largest 
loads  at  or  near  the  given  section. 

22.   Live  Load  Stresses  in  Trusses  and  Plate  Girders. 

Either  a  truss  or  a  girder,  under  the  action  of  live  loads, 
is  essentially  but  a  beam,  and  as  such  is  subjected  to 
bending  moments  and  vertical  shear. 

In  a  truss  such  as  that  shown  in  Fig.  25,  the  primary 
function  of  the  web  members  —  as  the  vertical  4-5  and 
the  diagonal  4-7  —  is  to  resist  the  vertical  shear,  since, 
if  the  chords  be  parallel,  these  web  members  are  the 
only  ones  having  either  vertical  stresses  or  stresses 
which  can  be  resolved  vertically.  Hence,  the  vertical 
component  of  the  stress  in  any  diagonal  of  this  truss  is 
equal  in  magnitude  to  the  vertical  shear  in  the  panel  in 
which  the  diagonal  is  located,  and  the  stress  in  the 
vertical  meeting  a  diagonal  at  a  panel-point  in  the 
unloaded  chord  is  equal  in  magnitude  and  opposite  in 
direction  to  the  vertical  component  of  the  stress  in  the 
diagonal. 

The  chord  members  —  as  4-6  and  5-7,  Fig.  25 — 
take  the  tensile  and  compressive  stresses,  respectively, 


88 


GRAPHIC   STATICS 


produced  by  the  bending  moments.  The  stress  in  either 
chord,  at  any  given  section  of  such  a  truss,  is  equal  to 
the  quotient  of  the  bending  moment  at  that  section, 
divided  by  the  depth  of  the  truss.  Thus,  Fig.  27  repre- 
sents a  segment  of  a  Pratt  truss  and  a  corresponding 

section  of  the  equilibrium 
polygon  constructed  for 
z  the  loads  and  reactions,  as 
for  a  simple  beam.  Ap- 
plying Rankine's  Method 
of  Sections,  let  the  truss 
be  cut  on  the  line  ab,  im- 
mediately to  the  right  of 
the  vertical  2-3 1  so  that 
the  bending  moments  at 
ab  and  2-3  will  be  virtu- 
ally the  same.  To  main- 
tain equilibrium,  apply  to  the  severed  members,  the  forces 
F,  Fv  and  F^  each  of  which  is  equal  to  and  like  the  stress 
in  the  member  to  which  it  is  applied.  The  segment  of 
the  truss  is  then  in  equilibrium  under  the  action  of  these 
applied  forces,  the  load  P  at  panel-point  j,  and  the  left 
reaction  Rv  The  two  latter  can  be  replaced  (Art.  11) 
by  their  resultant  r,  acting  vertically  through  the  point 
E,  the  intersection  of  the  prolonged  sides  BC  and  AD  of 
the  equilibrium  polygon.  For  equilibrium  to  exist,  the 
algebraic  sum  of  the  moments  of  r  and  the  three  applied 
forces,  about  an  axis  perpendicular  to  their  plane  of  action, 
must  be  zero.  Let  d  be  the  depth  of  the  truss.  Then, 
taking  moments  about  panel-point  2,  where  the  lines  of 


LIVE  LOADS:   SHEARS  AND  MOMENTS  89 

action  of  Fl  and  F%  meet  and  their  moments  are  zero, 
we  have : 

Fx  d—  r  X  / 

F  =  rl/d  =  Mjd. 

since,  by  Art.  1 1,  r  x  /  is  equal  to  the  bending  moment 
at  the  section  ab  or  2-3.  It  will  be  observed  that  these 
principles  relating  to  the  web-  and  chord-stresses,  apply 
only  to  trusses  with  parallel  chords.  When  the  upper 
chord  is  ' broken '  and  thus  inclined  to  the  lower,  there  is  a 
vertical  component  of  its  stress  which  must  be  considered. 
In  a  plate  girder,  the  web  plate  is  usually  assumed  to 
take  only  the  same  stresses  as  the  web  members  of  a 
truss  with  parallel  chords,  and  the  flanges  —  composed 
of  angles  and  cover-plates,  if  any  —  to  act  as  the  chord 
members.  The  customary  method  of  design  regards 
the  web  plate  as  resisting  vertical  shear  only  and  the 
flanges  as  taking  the  full  bending  stress ;  in  more  accu- 
rate work,  the  resistance  of  the  web  plate  to  bending  is 
considered. 

EXAMPLES 

15.  PLATE  GIRDER  BRIDGE  WITH  LOCOMOTIVE  WHEEL 
LOADS  :  MAXIMUM  MOMENTS  AND  SHEARS.  The  weight 
of  a  locomotive,  as  distributed  among  its  driving  and 
pilot  wheels,  forms  a  system  of  vertical  loads,  more  or 
less  unequal  and  unequally  spaced.  In  designing  a 
bridge,  the  maximum  moments  and  shears  must  be  de- 
termined for  the  greatest  live  load  which  will  pass  over 
it.  For  simplicity  in  the  following  example,  the  loco- 
motive only  will  be  considered ;  it  is  usual  to  design  the 


9o 


GRAPHIC    STATICS 


bridge  for  two  coupled  locomotives  with  their  tenders, 
followed  by  a  uniform  train  load. 

Let  AB,  Fig.  28,  represent  a  deck,  plate  girder  bridge 
for  a  single  track  railroad,  the  effective  span  and  depth 


of  the  girders  being  60  feet  and  6  feet,  respectively. 
The  load  carried  by  the  bridge  is  one  Northern  Pacific 
tandem  compound  locomotive  (1901),  whose  total  weight 
is  198,000  pounds,  175,000  pounds  being  on  the  driving 
wheels  and  23,000  pounds  on  the  engine  truck.  These 
weights,  when  distributed  as  shown  in  the  figure,  are,  in 


LIVE  LOADS:   SHEARS  AND  MOMENTS  91 

tons  of  2000  pounds,  10.94  tons  on  each  driver  and  5.75 
tons  on  each  pilot  wheel.  The  weights  on  the  wheels  at 
one  side  of  the  locomotive  constitute  the  system  of  live 
loads  borne  by  each  of  the  two  girders  forming  the 
bridge.  The  spacing  of  these  loads  is  marked  in  inches 
below  the  line  A'B'  indicating  the  span. 

(a)  Maximum  Moments.  The  system  of  loads  may 
be  located  in  any  position  on  the  span,  although  prefer- 
ably the  first  driving  wheel  should  be  near  the  middle  of 
the  girder,  as  in  the  figure.  The  moments  and  shears 
should  be  found  at  sections  reasonably  close  together 
throughout  the  span.  Therefore,  divide  the  girder  into 
any  number  of  segments,  say  10,  at  intervals  of  6  feet. 

On  the  load  line  win,  lay  off  the  loads  and  draw  the 
force  polygon  Omn,  with  pole  .distance  equal  to  5 
times  the  depth,  or  30  feet,  since,  by  using  thus  a  multi- 
ple of  the  depth,  the  moment  diagram  is,  by  Art.  22, 
available  for  the  measurement  of  flange  stresses.  Con- 
struct the  corresponding  equilibrium  polygon  acdefgb. 
Then,  the  bending  moment  at  any  one  of  the  numbered 
sections  of  the  girder  will  be  given  by  the  length  of  the 
ordinate  immediately  below  it  and  between  the  closing 
line  ab  and  the  perimeter  a  •••  e  •••  b  of  the  equilibrium 
polygon,  this  length  being  measured  by  the  proper 
moment  scale  (Art.  12). 

Now,  shift  the  span  one  interval  to  the  right,  as  at 
a$\\  prolong  the  side  gb\  project  a^  and  bl  to  a^  and 
'&i-,  respectively,  and  draw  the  new  closing  line  a^b^. 
This  operation  is,  in  effect,  moving  the  system  of  loads 
a  distance  equal  to  one  interval  to  the  left  on  the  span 


92  GRAPHIC    STATICS 

as  originally  placed,  and  the  equilibrium  polygon  for 
this  new  location  is  a^  •  •  •  e  •  •  •  b^ .  The  bending  moment 
at  any  section  of  the  girder  is  then  given  by  the  length 
of  the  corresponding  ordinate  above  that  section  (as 
marked  on  the  line  a-fi^  and  included  between  the 
perimeter  and  the  closing  line  a^b^. 

The  span  is  then  shifted  successively  to  several  new 
positions  to  the  left,  as  a2l?2,  etc.,  each  position  differing 
from  the  one  immediately  preceding  it  by  the  space  of 
one  interval ;  and  the  corresponding  equilibrium  poly- 
gon is  constructed  in  each  case.  This  process  is  con- 
tinued until  every  section,  from  the  middle  of  the  girder 
to  the  left  abutment,  has  been  subjected  to  its  maximum 
bending  moment.  Each  numbered  section  from  I  to  5, 
in  the  various  locations  of  the  span,  is  then  projected 
upon  the  corresponding  closing  line  of  the  equilibrium 
polygon,  and  a  free  curve,  known  as  the  stress  curve, 
is  drawn  through  the  points  thus  determined.  Thus, 
CD  and  EF  are  the  stress  curves  for  sections  i  and  2, 
respectively.  The  maximum  bending  moment  at  section 
I  is  evidently  given  by  the  length  of  the  greatest  ordi- 
nate included  between  the  stress  curve  CD  and  the  per- 
imeter a  •••  e  •  -•  b.  The  method  for  all  other  sections  is 
similar. 

The  maximum  moments  thus  obtained  for  the  live 
load  are  plotted  upward  from  the  axis  GK,  for  sections 
i  to  5,  giving  points  on  the  semi-curve  of  maximum 
moments  GL  ;  the  full  curve  GLK\s  symmetrical  about 
its  centre  line  LM,  and  is  cusped  at  L,  as  shown. 

The  dead  load,  i.e.,  the  total  weight  of  one  girder,  of 


LIVE  LOADS:   SHEARS  AND  MOMENTS  93 

one-half  the  lateral  bracing  and  cross-framing,  and  of 
the  floor  system,  may  be  taken,  in  this  case,  roughly,  as 
about  600  pounds  per  linear  foot  of  the  span,  or  18 
tons  for  each  girder.  This  load  is  considered  to  be  uni- 
formly distributed- over  the  span,  and  the  moments  due 
to  it  are  represented  by  the  curve  GNK,  which  is 
drawn,  for  convenience,  below  the  axis  G K,  to  the  same 
moment  scale  as  before.  The  maximum  moment  at  any 
section,  due  to  the  dead  and  live  loads,  will  then  be 
given  by  the  length  of  the  corresponding  ordinate 
included  between  the  two  curves.  The  curve  GNK  is 
a  parabola  whose  equation  (Art.  n)  is  : 

M=  WSX/2  —  ZVX^/2, 

in  which  M  is  the  bending  moment  in  ton-feet  at  the 
section  distant  x  feet  from  the  left  support,  w  is  the 
uniform  load  in  tons  per  lineal  foot  of  the  span,  and  s  is 
the  span  in  feet. 

(b)  Maximum  Shears.  It  can  be  shown  by  compu- 
tation or  by  measurement  from  the  diagrams  that  the 
maximum  live  load  shear  at  any  section  occurs  when  the 
first  driving  wheel  is  immediately  above  that  section,  the 
shear  being  then  equal  to  the  left  reaction,  less  the  load 
of  5.75  tons  on  the  pilot  wheel.  The  maximum  shear 
can  hence  be  found  from  a  diagram  similar  to  that  in 
Fig.  28,  by  successively  locating  each  section  of  the 
span  below  the  first  driver,  constructing  the  correspond- 
ing equilibrium  polygon,  and  then  (Art.  n)  drawing  a 
ray  in  the  force  polygon  parallel  to  the 'closing  line  of 
the  equilibrium  polygon.  The  distance  between  the  in- 


94 


GRAPHIC    STATICS 


tersection  of  this  ray  with  the  load  line  mn  and  the  point 
m  will  then  be  the  left  reaction  for  that  position  of  the 
loads,  and,  deducting  from  this  the  first  load  of  5.75  tons 


FIG.  29 

on  the  pilot  wheel,  we  have  the  maximum  live  load  shear 
at  the  section  immediately  under  the  first  driving  wheel. 
A  more  convenient  method  is  shown  in  Fig  29.  A 
new  force  polygon  O'm'n'  is  constructed,  having  the  pole 
O'  at  a  horizontal  distance  from  m1  equal  to  the  span. 


LIVE  LOADS:   SHEARS  AND  MOMENTS  95 

Then,  assuming  the  loads  to  be  located  as  on  the  span 
A'B' ,  Fig.  28,  the  corresponding  equilibrium  polygon  will 
be  a'c'd'e'fg'b1 .  In  the  force  polygon,  draw  the  ray 
O'a"  parallel  to  the  closing  line  b'd.  Since  the  tri- 
angles O'm'a"  and  b"a'b'  are  equal  in  all  respects,  b'bn  is 
equal  to  a"m' ,  and  hence  to  the  left  reaction  for  this 
position  of  the  loads.  This  principle  is  general,  applying 
to  all  other  values  of  the  left  reaction,  when  found 
similarly. 

Prolong  the  sides  g'b'  and  a 'c',  and  draw  the  indefinite 
line/^  parallel  to  d c\  and  at  a  vertical  distance  from  it 
equal  to  the  pilot  load,  5.75  tons.  Lay  out  the  span 
with  section  5  immediately  below  the  first  driving 
wheel ;  the  ordinary  y  above  the  right  end  of  the  span  is 
the  left  reaction  for  this  position  of  the  loads,  and  the 
ordinate  j/5  is  the  maximum  live  load  shear  at  section  5 
since  the  pilot  load  must  be  deducted.  Similarly,  j/3  is 
the  similar  shear  at  section  3,  and  so  on.  When  the 
first  driver  reaches  the  left  support,  its  line  of  action  co- 
incides with  that  of  the  left  reaction,  and  the  pilot  load 
is  no  longer  on  the  girder.  Hence,  the  side  c'd'  is  pro- 
longed to  d",  and  the  live  load  shear  at  section  o  is 
equal  to  the  left  reaction,  and  is  represented  by  the 
ordinate  j0.  In  this  method,  it  is  assumed  that  the  given 
sections  —  as  5,  j,  and  o  —  are  successively  located  im- 
mediately below  the  first  driving  wheel,  which  is  only 
approximately  the  position  shown  in  Fig.  29,  the  latter 
position  being,  for  comparison,  the  same  as  that  in  Fig. 
28. 

As  stated  previously,  the  live  load  shear  is  positive 


96  GRAPHIC    STATICS 

when  the  load  moves  from  right  to  left.  Laying  off 
these  shears  from  the  axis  PQ  upward,  PQRS  becomes 
the  diagram  of  live  load  shears  from  the  left  support  to 
the  middle  of  the  girder.  The  total  dead  load  on  each 
girder  is  18  tons.  As  this  load  is  uniformly  distributed, 
the  dead  load  shear  is  one-half  of  it,  or  9  tons,  at  the  left 
support,  and  decreases  uniformly  to  zero  at  the  middle 
(Art.  8).  This  shear  is  also  positive,  but,  for  conven- 
ience, is  laid  off  below  the  axis  in  the  diagram  TPQ. 
The  final  shear  at  any  section  is  then  represented  by 
the  corresponding  ordinate  in  the  combined  diagram 
STQR.  The  diagram  for  the  right  half  of  the  girder, 
and  with  the  load  moving  from  left  to  right,  will  be 
similar  but  reversed,  corresponding  sections  having 
shears  of  the  same  magnitude,  but  both  those  due  to 
dead  and  live  loads  being  negative. 

(c)  Flange  Stresses:  Impact.  By  Art.  n,  the  bend- 
ing moment  at  any  section  of  a  beam  is  equal  to  the 
product  of  the  corresponding  ordinate  in  the  equilibrium 
polygon  by  the  pole  distance  H  of  the  force  polygon. 
By  Art  22,  the  flange  stress,  which  corresponds  with  the 
chord  stress  in  a  truss,  is  equal  to  the  bending  moment, 
divided  by  the  depth  of  the  truss  or  girder.  In  this 
case,  the  pole  distance  in  the  first  force  polygon  was 
taken  as  5  times  this  depth.  Hence  : 

M=y  x  H, 

77=54 

A  x  S  =  M/d=  Hy/d=  57, 

in  which  M  is  the  bending  moment  at  the  given  section, 
y  is  the  corresponding  ordinate  of  the  bending  moment 
diagram,  p  is  the  depth  of  the  girder,  A  is  the  total  area 


LIVE  LOADS:  SHEARS  AND  MOMENTS 


97 


of  either  flange,  and  5  is  the  flange  stress,  per  square 
inch,  at  the  given  section.  Hence,  the  moment  diagram 
GNKL,  Fig.  28,  will  serve  as  a  total  (A  x  5)  flange 
stress  diagram,  if  its  ordinates  be  measured  by  a  force 
scale  5  times  greater  than  that  used  for  the  force  poly- 
gon, i.e.,  if  the  scale  of  Omn  be  20  tons  to  the  inch,  the 
flange  stress  scale  will  be  100  tons  to  the  inch.  The 
unit  stress  at  any  given  section  of  the  web,  due  to  the 
vertical  shear,  can  be  found  by  dividing  the  total  shear, 
as  given  by  the  corresponding  ordinate  of  the  diagram 
STQR,  Fig.  29,  by  the  cross-sectional  area  of  the  web. 

In  bridges,  the  stresses,  as  determined  above,  are  in- 
creased by  an  allowance  for  the  impact  resulting  from 
the  sudden  application  of  the  live  load.  Formulas  for 
this  allowance  are  given  in  works  on  this  subject;  the 
increase  is  usually  about  70  per  cent. 

(d)  Summary.  Tabulating  the  results,  as  measured 
from  the  diagrams  by  the  moment  scale,  we  have,  neg- 
lecting the  allowance  for  impact : 


SECTION 

0 

I 

1              2 

3              4 

5 

Bending  moment, 

i 

dead  load,  ton-feet         o     I  •  48.6  !    86.4  i  113.4  \  129.6     135. 

Maximum  moment, 

live  load,  ton-feet    .        o 

198.3  1  354-    1  498.5 

558. 

588.5 

Total  bending  moment      o       246.9 

440.4   !   6II.9 

687.6 

723.5 

Vertical  shear,  dead 

load,  tons      ...        9 

7.2 

5'4 

3.6        1.8 

0 

Maximum  live  load 

shear,  tons    . 

37-5 

33.2 

28.5 

23-5 

18.6 

13-7 

Vertical  shear,  total, 

tons      46.5 

40.4 

33-9 

27.1 

20.4 

13-7 

i           !                                      ' 

9»  GRAPHIC   STATICS 

1 6.  WARREN  GIRDER  FOR  OVERHEAD  CRANE  :  MAXI- 
MUM STRESSES.  Girders  for  overhead  cranes  are  of  sev- 
eral types.  The  rolled  beam,  the  plate  girder,  the  box 
girder,  and  the  truss  in  different  forms  have  each  its  place 
for  this  service.  The  selection  depends  mainly  on  the 
span  and  the  maximum  live  load  to  be  transported, 
although  economy  in  weight  and  construction  are  pri- 
mary requisites. 

The  upper  diagram,  Fig.  30,  is  the  elevation  in  out- 


FIG.  30 

line  of  a  Warren  girder  or  truss  with  parallel  chords, 
as  employed  in  such  cranes.  In  this  girder,  the  web 
members  are  usually  all  diagonals  and  they  are  built  to 
withstand  both  tension  and  compression ;  verticals  may 
be  inserted  as  struts  between  each  joint  of  the  lower 
chord  and  the  middle  of  the  panel  above  it,  when  the 
length  of  a  panel  is  considerable.  The  crab,  which  car- 
ries the  live  load,  has  four  wheels,  tracking  on  the  upper 


LIVE  LOADS:  SHEARS  AND  MOMENTS  99 

chords  of  the  two  parallel  girders.  The  conditions  are 
thus,  in  general,  the  same  as  those  of  Example  15,  ex- 
cept that  the  plate  girder  has  been  replaced,  in  this 
case,  by  the  Warren  truss.  Let  the  girder,  Fig.  30,  be 
a  6-panel,  deck  truss  of  72  feet  span  and  6  feet  in  depth. 
Consider  first : 

(a)  Dead  Load  Stresses.  The  dead  load  on  each 
girder  consists  of  the  weight  of  one  girder,  its  cross- 
shafting,  and  platform.  Take  it  as  6  tons  and  consider 
it  as  uniformly  distributed  over  the  girder.  Then,  as  in 
the  roof  truss,  Example  7,  there  will  be  a  weight  W 
=  i  ton,  acting  at  each  of  the  inner  joints  of  the  upper 
chord,  and  a  weight  W ' /2  =  %  ton,  on  each  of  the  end 
joints.  Since  the  load  is  uniformly  distributed,  the  re- 
actions R1  and  R2  are  equal  and  each  is  3  tons. 

Lay  off  the  loads  ab,  be,  etc.,  on  the  load-line  aaf. 
Bisect  this  line  at  m.  Then,  ma  —  R1  and  a'm  =  R2. 
The  closed  force  polygon  for  the  loads  and  reactions  is 
then  abcdd' c' b' a1  ma.  From  m,  draw  the  indefinite  hori- 
zontal line  ml. 

Joint  I  is  in  equilibrium  under  the  action  of  the  reac- 
tion Rl  and  the  load  W/2t.  which  are  known,  and  the 
stresses  in  the  members  BE  and  EM,  which  stresses 
are  to  be  determined.  From  b  and  m  lay  off  be  and 
me,  parallel  respectively  to  BE  and  EM  and  meeting 
at  e.  Then,  taking  the  loads  in  clockwise  order,  the 
closed  force  polygon  for  the  joint  is  abema.  The  stress 
be  thus  acts  toward  the  joint  and  is  therefore  compres- 
sive,  while  em  acts  from  it  and  is  tensile. 

At  joint  2  there  are  acting  the  known  stress  me  and 


100 


GRAPHIC    STATICS 


the  stresses  in  EFand  FM,  which  are  to  be  determined. 
From  e  and  m,  lay  off  ef  and  mft  parallel  to  their  cor- 
responding members  and  meeting  at/.  The  force  poly- 
gon is  mefm,  efis  compressive,  and/;;/  is  tensile. 

In  a  similar  way,  the  force  polygon  for  joints  j  to  8, 
inclusive,  are  drawn.  Thus,  for  joint  j,  the  polygon  is 
bcgfeb\  for  joint  4,  fghmf\  etc.  Since  the  load  is  uni- 
formly distributed,  the  stress  diagram  is  symmetrical 
about  its  axis  Im.  It  will  be  seen  that  the  diagonals. 


when  under  dead  load,  are  alternately  struts  and  ties, 
the  order  changing  at  the  middle  of  the  truss. 

(b)  Live  Load  Stresses.  The  maximum  weight  to  be 
transported  may  be  taken  as  15  tons  and  the  weight  of 
the  crab  as  5  tons,  which  will  bring  a  load  of  20/4  =  5 
tons  on  each  of  the  four  wheels  or  runners  of  the  crab. 
The  distance  between  these  runners,  measured  parallel 
to  the  girder,  is  6  feet. 

The  graphic  analysis  of  a  truss  is  based  on  the  as- 
sumption (Art.  5)  that  all  loads  are  transferred  to,  and 


LIVE  LOADS:  SHEARS  AND  MOMENTS  101 

concentrated  at,  the  joints  of  the  truss.  Hence,  if  the 
centre  of  the  crab  be  above  joint  j, —  as  in  the  partial 
diagram,  Fig.  31,  —  each  runner  will  be  3  feet  from  that 
joint  and  9  feet  from  the  joint  on  the  other  side  of  the 
runner.  The  load  of  5  tons  on  each  runner  will  then 
be  divided  in  the  proportion  of  3  to  I  between  the  two 
adjacent  joints.  In  the  position  shown,  there  will  thus 
be  a  load  P  =  3.75  x  2  =  7.5  tons  on  joint  3,  a  load  Pl  = 
1.25  tons  on  joint  i,  and  a  load  P2  =  1.25  tons  on  joint  5. 
The  crab  and  the  weight  transported  thus  form  a  sym- 
metrical system  of  three  loads  spaced  12  feet  apart. 

The  maximum  stresses  produced  in  the  members  of 
the  truss  by  the  travel  of  the  crab  and  its  load  can  be 
found  by  locating  the  system  of  loads  centrally  above 
panel-points  j,  5,  and  7,  in  succession,  and  taking,  for 
each  member,  the  greatest  of  the  three  stresses  developed 
in  it  by  these  three  locations  of  the  system  of  loads. 
Since  the  crab  may  approach  the  middle  of  the  girder 
from  either  end,  it  will  be  necessary  to  determine  the 
stresses  in  one-half  only  of  the  members. 

There  are  several  ways  of  rinding  the  changes  in  the 
stresses  due  to  these  changes  in  the  location  of  the  loads  : 
first,  the  method  of  the  equilibrium  polygon  can  be  em- 
ployed, as  in  Example  1 5  ;  second,  the  system  can  be 
located  at  each  of  the  three  panel-points  in  succession, 
and  a  separate  stress  diagram  drawn  for  each  case ;  and, 
finally,  a  single  stress  diagram,  for  one  load  only  at  one 
panel-point,  can  be  constructed,  and  the  stresses  due 
to  the  system  at  each  of  the  three  panel-points  can  be 
found  by  proportion  from  this  stress  diagram.  This 


IO2 


GRAPHIC    STATICS 


method  of  multiples  is  essentially  the  use  of  the  influence 
diagram  in  modified  form.  While  possibly  not  the  most 
convenient  process  in  this  case,  it  is  given  below  for  in- 
formation. 

Assume  the  central  load  P=  7.5  tons  to  be  the  only 
one  on  the  girder  and  to  be  located,  as  in  Fig.  31,  at 
panel-point  j.  The  reactions  are  :  R1  =  6.2$  tons  and 
R<t=  1.2$  tons.  The  closed  force  polygon  for  the  loads 
and  reactions  is  bcmb.  The  stress  diagram  is  constructed 
by  the  same  method  as  that  used  for  the  dead  load  dia- 
gram. The  stresses  thus  determined  for  the  web  mem- 
bers are  tabulated  under  'panel-point  j'  in  Table  i. 

TABLE   i 


WEB  MEMBER 

CENTRAL  LIVE  LOAD  P  AT  PANEL-POINT 

3                       5 

7 

9 

ME                   +  8.9 

+  7-1            +5-3 

+  3-6 

EF 

-8.9 

-7-1 

-5-3 

-3.6 

FG 

-  1.8 

H-7-i 

+  5-3 

+  3-6 

GH 

+  1.8 

-7-1 

-5-3 

-3.6 

HK 

-  1.8 

-3-6 

+  5-3 

+  3-6 

KL 

4-  1.8 

+  3-6 

-5-3 

-3.6 

The  values  given  in  the  three  remaining  columns  of 
the  table  are  computed  from  those  in  the  first  column. 
The  method  is  as  follows : 

The  stress  in  a  diagonal  member  is  directly  propor- 
tional to  the  vertical  shear  in  the  panel  in  which  the 


LIVE  LOADS:  SHEARS  AND  MOMENTS 


103 


diagonal  is  located.  Since  there  is  but  one  load,  the 
vertical  shear  is  equal  to  the  left  reaction.  When 
the  load  P  is  at  panel-point  j,  R1  —  5/6  P  and  R%  —  1/6 
P  \  when  P  passes  to  panel-point  5,  Rl  =  4/6  P  and  R% 
=  2/6  P.  Hence,  in  the  latter  case,  the  stresses  in  all 
diagonals  to  the  left  of  P  are  4/5  of  those  in  the  simi- 
lar diagonals  in  the  former  case ;  while,  on  the  right  of 
P,  the  multiple  is  2  and  the  stresses  are  double  those 
which  prevailed  previously  in  similar  diagonals.  With 
P  at  panel-point  7,  the  multiples  to  left  and  right  of  P 
are,  respectively,  3/5  and  3 ;  and  so  on.  The  signs  of 
the  stresses  are  determined  by  the  fact  that  the  two 
diagonals  meeting  at  the  panel-point  where  P  is  located 
are  under  compression,  and  that  thereafter,  both  to  left 
and  right,  the  diagonals  are  alternately  under  tension 
and  compression.  As  before,  tensile  stresses  are  con- 
sidered as  positive  (  +  )  and  compressive  stresses  as 

TABLE   2 


WEB  MEMBER 

END  LOAD  Pj  OR  Pz  AT  PANEL-POINT 

3 

5 

7 

9 

ME 

+  i-5 

-f  1.2 

•f  0.9 

+  0.6 

EF 

-  i-S 

y      2 

-0.9 

-  0.6 

FG 

-o-3 

+  1.2 

4-  0.9 

+  0.6 

GH 

+  0.3 

—  1.2 

-0.9 

-0.6 

HK 

-0.3 

-0.6 

4-  0.9 

+  0.6 

KL 

+  0.3 

+  0.6 

-  0.9 

-0.6 

104 


GRAPHIC    STATICS 


negative  ( —  ).  The  values  above  the  heavy  line  in  the 
table  refer  to  members  to  the  left  of  the  load  P. 

The  load  /^  = />2  =  1.25  tons  is  equal  to  P/6. 
Hence,  in  the  same  location  as  P,  it  will  develop 
stresses  of  the  same  kind  and  only  one-sixth  as  large. 
Dividing  the  values  in  Table  I  by  6,  we  have  the 
stresses  in  Table  2  for  either  Pl  or  P2,  as  it  moves 
across  the  girder. 

When  the  system  of  three  loads  is  centrally  above 
a  panel-point,  P  is  at  the  latter  point  and  P1  and  P2 
are  at  adjacent  panel-points  to  the  left  and  right,  re- 
spectively. Selecting  the  values  for  these  conditions 
from  Tables  i  and  2,  we  have,  in  Table  3,  the  live  load 

TABLE   3 


WEB  MEMBERS 

ME 

EF 

FG 

G7/ 

HK 

KL 

SYSTEM  AT  PANEL-POINT  j: 

Load  PI  at  panel-point   /.     .         o 

o 

o 

o 

o 

•o 

Load  P  at  panel-point  j.     .1+8.9 

-   8.9  I  —  1.8    +1.8    —1.8 

+  1.8 

Load  />2  at  panel-point  5  .     .  j  +    i  2 

-    1.2  ;  +  1.2 

—  1.2     —0.6 

+  1.6 

Stresses  due  to  system     .     .     .  '  +  10.1 

—  10.1    —0.6    +0.6 

—  2.4 

+  2.4 

i 

SYSTEM  AT  PANEL-POINT  5  :   j 

Load  PI  at  panel-point  j  .     .      +1.5 

—  1.5     —0.3    +0.3 

-0.3    +0.3 

Load  P  at  panel-point  5  .     .      +7.1 

-7.1 

+  7-1    -7-1 

-3.6I+3.6 

Load  /2  at  panel-point  7  .     .+09 

—  0.9 

+  0.9'—  0.9   +0.9—0.9 

Stresses  due  to  system     .     .     .      +9-5 

—  9-5     +7-7;—  7,7    —3-    |+3- 

SYSTEM  AT  PANEL-POINT  7  :  '. 

Load  PI  at  panel-point  5  .     . 

+  1.2 

—  1.2 

+  1.2     —1.2 

—  0.6+0.6 

Load  P  at  panel-point  7  .     . 

+  5-3 

-5-3 

+  5-3 

-  5-3 

+  5-3-5-3 

Load  />2  at  panel-point  g  .     . 

+  0.6 

—  0.6 

+  0.6 

—  0.6 

+  0.6    —0.6 

Stresses  due  to  system     .     .     . 

+  7-1 

—  7.1 

+  7-1 

-7.1 

+  5-3 

-5-3 

LIVE  LOADS:  SHEARS  AND  MOMENTS 


105 


stresses  in   the  web   members   for  the   three   required 
positions  of  the  system  of  loads. 

Taking  the  maximum  and  minimum  live  load  stresses 
from  Table  3  and  combining  them  with  the  dead  load 
stresses,  as  measured  from  the  stress  diagram,  Fig. 
30,  we  have,  in  Table  4,  the  maximum  stresses  in  the 

TABLE   4 


WEB  MEMBERS 

STRESS  RANGE, 
LIVE  LOAD 

DEAD  LOAD 

STRESSES 

MAXIMUM 
STRESSES 

ME 

+  7.1  to  +  10.  i 

+  3-6 

+  13-7 

EF 

-  7.1  to  —  10.  i 

-3-6 

-  13-7 

FG 

—  0.6  to  -f    7.7 

+  2.2 

+    9-9 

GH 

+  0.6  to  —    7.7 

—  2.2 

-    9-9 

HK 

-  2.4  to  H-    5.3 

+  0.7 

+    6. 

KL 

'+  2.4  to  -     5.3 

-  0-7 

-    6. 

web    members,    as    the    crab,    carrying    its    live   load, 
travels  from  the  end  to  the  middle  of  the  girder. 

17.  PRATT  TRUSS  :  UNIFORM  LIVE  LOAD.  The  mag- 
nitudes of  the  stresses  in  the  members  of  a  bridge 
truss,  due  to  dead  loads  and  to  uniform  live  loads,  can 
be  determined  by  stress  diagrams,  as  with  a  roof  truss 
and  with  the  Warren  truss,  Fig.  30;  these  stresses 
can  be  found  also  by  the  general  method  of  the  force 
and  equilibrium  polygon.  The  use  of  the  latter  is 
illustrated  in  Fig.  32,  which  represents  a  Pratt  truss, 
through  type,  of  six  panels,  having  a  span  of  144  feet 
and  a  depth  of  24  feet.  Let  the  uniform  live  load 


io6 


GRAPHIC    STATICS 


crossing  the  bridge  be  a  train  weighing  4000  pounds 
per  lineal  foot.  One-half  of  this  load  will  be  carried 
by  each  of  the  two  trusses  forming  a  single-track  rail- 
way, as  in  this  case.  This  gives  a  load  of  4000/2 
=  2000  pounds  =  i  ton  per  lineal  foot  of  span,  or  a 
panel  load  of  24  tons. 


FIG.  32. 

Each  of  the  inner  panel-points,  3,  5,  7,  p,  and  //, 
carries  this  load ;  the  end  points,  i  and  jj,  have  each 
a  half  panel  load,  or  12  tons.  As  the  total  load  is  uni- 
formly distributed,  the  two  reactions  will  be  equal. 
The  total  reaction  at  the  left  support  is  24  x  6/2  =  72 
tons;  deducting  the  half  panel  load  at  point  i,  the 
effective  reaction  is  60  tons. 

To  any  convenient  scale,  lay  off  the  loads  and  effec- 
tive reactions  on  the  load  line  bb\  and,  taking  the  pole 
distance  H  as  a  multiple  of  the  depth  of  the  truss,  draw 
the  force  polygon  Obb'  and  the  corresponding  equilib- 


LIVE  LOADS:  SHEARS  AND  MOMENTS  107 

rium    polygon    KLM,  which  polygon  is   the   bending 
moment  diagram  for  the  truss. 

Since  the  vertical  shear  V  is  equal  to  the  left  reac- 
tion, minus  any  loads  to  the  left  of  the  section  con- 
sidered, the  shear  diagram  efg,  with  zero  line  kkt  can 
be  drawn,  as  shown,  by  projection  from  the  lower  chord 
of  the  truss  and  from  the  load  line  of  the  force  polygon. 
Thus,  in  the  first  panel,  V=Rl  =  ab\  in  the  second 
panel,  V—  R^  —  load  BC  =  ab  —  be  =  ac ;  etc. 

(a)  Chord  Stresses.     By  Art.  22,  the  total  stress  in 
either  chord  at  any  given  section  of  the  truss  is  equal 
to  the  bending  moment  at  that  section,  divided  by  the 
depth  of  the  truss.     In  this  case,   as  in  Example   15, 
the  pole  distance  is  a  multiple  of  the  depth  of  the  truss. 
Hence,  by  using  a  stress  scale  proportioned  for  these 
conditions,  the  chord  stress  at  any  given   section    can 
be  measured  directly  from  the  moment  diagram  KLM. 
Thus,  at  the  middle  of  the  second  panel,  the  total  stress 
in  either  chord  is  78  tons;  at  the  middle  of  the  third 
panel,  102  tons ;  etc.     These  stresses  are  compressive 
in  the  upper  chord  and  tensile  in  the  lower. 

(b)  Stresses  in  Diagonals.     By  Art.  22,  the  vertical 
component  of  the  stress  in  any  diagonal  member  of  a 
truss  with  parallel  chords  is  equal  to  the  vertical  shear 
in  the  panel  in  which  the  diagonal  is  located.     Thus, 
a"c"  =  36  tons  is  the  vertical  shear  in  the  second  panel 
and    is    also    the  vertical    component    of    c"a",    drawn 
parallel    to    the   diagonal   2-5.      The   length    of   c"d", 
measured  by  the  same  scale,  is  51.4  tons,  which  is  the 
stress  in  the  diagonal  2-5. 


108  GRAPHIC    STATICS 

(c)  Stresses  in  Verticals.     By  Art.  22,  the  stress  in  the 
vertical  member  of  a  truss  with  parallel  chords  is  equal 
to  the  vertical  component  of  the  stress  in  the  diagonal 
which  meets  the  vertical  at  a  panel-point  in  the  unloaded 
chord,  and  is  hence  equal  to  the  vertical  shear  in  the 
panel  in  which  that  diagonal  is  located.    Thus,  the  stress 
in  the  vertical  at  panel-point  j  is  equal  to  the  vertical 
shear  in  the  second  panel,  or  36  tons,  etc.     The  stress 
in  the  central  vertical  at  panel-point  7  is  equal  to  the 
algebraic  sum  of  the  shears  in  the  two  adjacent  panels, 
or  zero  in  this  case. 

(d)  Character  of  the  Stresses.     The  character  of  the 
stresses,  as  to  tension  or  compression,  cannot  be  deter- 
mined from  the  equilibrium  polygon,  but  must  be  ascer- 
tained from  other  considerations.     Thus,  in  any  truss 
the  upper   chord   is  in   compression  and  the  lower  in 
tension.     Again,  the  positive  or  negative  character  of 
the  vertical  shear  in  any  panel  is  known,  and,  using  the 
method  shown  in    Fig.  25,  the  nature  of   the  stresses 
in  the  diagonals  and  verticals,  respectively  in  and  adja- 
cent to  that  panel,  can  be  found.     In  the  Pratt  truss, 
the  stresses  in  all  verticals  are  compressive  and  those 
in    all    diagonals  in   action    are   tensile ;    in  the  Howe 
truss,  the  main  diagonals  have  the  opposite  inclination, 
and,    as    the   character    of    the   vertical    shear   is    the 
same,  the  stresses  in  the  web  members  are  the  reverse 
of  those  in  the  Pratt  truss.     In   both    trusses,  one  or 
more  panels  at  the  middle  are  counterbraced  (Art.  19) 
to  provide  for  the  reversal  of  shear  when  concentrated 
loads  cross  the  bridge. 


LIVE  LOADS:  SHEARS  AND  MOMENTS  109 

(e)  Concentrated  Loads.  The  graphic  analysis  of  the 
live  load  stresses,  due  to  a  series  of  concentrated  loads, 
in  this  and  similar  trusses,  can  be  made  most  conven- 
iently by  the  use  of  the  force  and  equilibrium  polygons, 
as  in  Example  15.  The  method,  while  essentially  simple, 
would  require  more  space  than  is  available  here,  for 
description  in  detail.  In  outline  it  is  as  follows  : 

For  convenience  in  investigation,  a  tentative  truss 
diagram  is  drawn  with  the  diagonals  all  inclined  in  one 
way  —  that  of  the  members,  2-5  and  4-7,  Fig.  32. 
This  uniform  inclination  does  not  alter  the  magnitude 
of  the  stress  in  any  diagonal ;  it  simply  changes,  from 
tension  to  compression,  the  stresses  in  those  members 
which  will  have  the  opposite  inclination  in  the  truss  as 
built.  Let  the  series  of  concentrated  loads  be  those  of 
two  coupled  locomotives  followed  by  a  train,  all  moving 
from  right  to  left.  Proceeding  as  in  Example  15,  the 
stress  curves  are  drawn  and  the  maximum  shear  found, 
both  for  half-panel  intervals  from,  the  left  abutment  to 
the  middle  of  the  truss.  From  the  stress  curves,  the 
diagram  for  maximum  bending  moments  and  chord 
stress  can  be  drawn ;  from  the  maximum  shears,  the 
diagram  for  the  latter  can  be  constructed  and  the  stresses 
in  the  diagonals  determined,  as  in  the  shear  diagram, 
Fig.  32.  The  stresses  in  the  verticals  are,  in  general, 
equal  to  the  vertical  shear  in  the  panel  to  the  right  of  it 

The  stresses,  as  thus  found,  are  tabulated  with  those 
due  to  dead  load,  in  order  to  determine  the  maximum 
stress  in  each  member  and  to  ascertain  which  panels 
require  counterbracing,  owing  to  a  reversal  of  stress  in 
the  diagonals. 


CHAPTER   V 

CENTRE  OF  GRAVITY  :    MOMENT  OF  INERTIA 

In  engineering  calculations,  the  location  of  the  centre 
of  gravity  of  a  plane  area  is  frequently  required;  for 
example,  through  this  point  the  neutral  axis  of  the 
cross-section  of  a  beam  passes.  Similarly,  the  magni- 
tude of  the  rectangular  moment  of  inertia,  about  an 
axis  through  the  centre  of  gravity  of  the  cross-section, 
as  above,  must  be  known,  in  order  to  determine  the 
stresses  due  to  bending  at  that  section  of  the  beam. 
The  centre  of  gravity  and  the  moment  of  inertia  of 
plane  areas  can  both  be  found  graphically  by  the 
methods  given  below. 

23.  Centre  of  Gravity.  The  weight  of  a  body  is  the 
result  of  the  action  of  terrestrial  gravitation  upon  the 
latter.  A  body  is  composed  of  an  indefinite  number 
of  indefinitely  small  particles;  the  earth  attracts  each 
of  the  latter  with  a  force  proportionate  to  the  mass  of 
the  particle.  These  forces  of  attraction  thus  form,  for 
each  body,  a  system  whose  components  are  vertical  and 
virtually  parallel,  since  the  centre  of  mass  of  the  earth 
is  4000  miles  distant  and  each  particle  is  indefinitely 
near  those  adjacent  to  it. 

The  resultant  force  of  this  system  is  the  weight  of  the 
body,  and  the  line  of  action  of  this  resultant  passes 

no 


CENTRE  OF  GRAVITY:  MOMENT  OF  INERTIA   III 

through  the  centre  of  gravity  or  mass-centre  of  the  body. 
In  treating  areas  which  have  no  mass,  it  is  assumed 
that  the  force  of  gravity  acting  on  the  area  is  propor- 
tional to  the  magnitude  of  the  latter,  i.e.,  that  there  is 


FIG.  33 


an  equal  force  applied  to  the  centre  of  each  unit  of  area, 
as  if  the  latter  were  that  of  a  thin  plate. 

24.  Centroid  of  Two  Parallel  Forces.  The  centre  of 
gravity  is  virtually  the  centroid  or  centre  of  a  system 
of  parallel  forces.  In  Fig.  33,  let  Pl  and  P2  be  two 
parallel  forces  acting  from  the  points  of  application  A 
and  B,  respectively.  Draw  the  force  polygon  Oac  and 


112  GRAPHIC    STATICS 

the  equilibrium  polygon  ADE\  the  resultant  R  —  Pl 
-f-  P2  acts  on  the  line  GD  dividing  the  line  AB  into 
segments,  AG  and  BG,  which  are  inversely  as  the 
forces  applied  at  A  and  B,  respectively.  That  GD  is 
the  line  of  action  of  R  follows  from  the  fact  that  R  =  ac, 
and  hence  the  intersection  of  the  sides  AD  and  ED, 
drawn  parallel  to  the  rays  Oa  and  Oc,  must  lie  at  D  on 
a  point  in  the  line  of  action  of  R\  this  determines  that 
line,  since  it  must  also  be  parallel  to  the  lines  of  action 
of  P1  and  P2. 

Now,  revolve  the  forces  P1  and  P2  about  their  points 
of  application,  forming  the  system  of  parallel  forces  P± 
and  /y,  whose  force  and  equilibrium  polygons  are 
Ofa'cf  and  A'D'E',  respectively.  The  line  of  action  of 
the  resultant  R'  of  this  new  system  acts  on  the  line 
GD1  which,  as  before,  intersects  AB  at  G.  With  the 
same  relative  locations  of  the  poles  O  and  0',  the  con- 
struction of  the  force  polygon  O'a'c'  is  unnecessary, 
since  the  sides  of  the  equilibrium  polygon  A'D'E'  are 
at  the  same  angle  with  the  corresponding  sides  of  the 
polygon  ADE  as  that  through  which  the  forces  were 
revolved. 

It  is  evident  that,  at  whatever  angle  the  forces  be 
inclined,  if  they  remain  parallel,  the  line  of  action  of 
their  resultant  will  pass  through  the  point  G,  which  is 
hence  the  point  of  application  of  that  resultant.  This 
point  is  called  the  centre  of  parallel  forces  or  the  centroid. 
In  Fig.  33,  the  forces  have  the  same  direction.  If  they 
act  in  opposite  directions  and  are  still  unequal  in  mag- 
nitude, their  resultant  will  be  equal  to  their  algebraic 


CENTRE    OF    GRAVITY:    MOMENT    OF    INERTIA        113 

sum  and  will  act  on  a  line  parallel  to  their  lines  of 
action  and  meeting  AB  prolonged  at  a  point  correspond- 
ing with  G  in  Fig.  33.  If  the  forces  are  equal  in  mag- 
nitude and  opposite  in  direction,  they  form  a  couple 
whose  centroid  is  infinitely  distant  from  the  points  of 
application  of  the  forces  and  on  a  line  drawn  through 
these  points. 

25.  Centroid  of  Complanar  Parallel  Forces  whose 
Points  of  Application  are  Complanar  with  all  of  the 
Forces,  but  are  not  in  a  Straight  Line.  These  are  the 
conditions  which  exist  in  the  graphical  determination  of 
the  centre  of  gravity  of  an  area.  It  is  evident  that,  so 
long  as  the  points  of  application  of  parallel  forces  are 
in  the  same  straight  line,  the  centroid  of  the  system  will 
lie  at  the  intersection  of  the  line  of  action  of  the  result- 
ant with  that  line,  whatever  may  be  the  number  or  rela- 
tive direction  of  the  forces.  When,  however,  the  points 
of  application,  although  in  the  same  plane,  do  not  lie  in 
the  same  straight  line,  the  method  shown  in  Fig.  33, 
i.e.,  the  intersection  of  the  lines  of  action  of  resultants, 
can  be  used  for  complanar  forces  in  determining  the 
centroid  of  the  system. 

In  Fig.  34,  let  P1  •••  P4  be  a  system  of  parallel  forces 
having  the  points  of  application  A,  B,  C,  and  D,  re- 
spectively, these  points  and  the  lines  of  action  being  in 
the  same  plane.  Draw  the  force  and  equilibrium  poly- 
gons, Oae  and  EBFLH,  respectively  ;  the  resultant  R 
acts  on  the  line  RHK.  Revolve  the  forces  about  their 
points  of  application,  keeping  the  lines  of  action  still  in 
their  original  plane.  The  system,  PJ  •••  /Y>  is  thus 


GRAPHIC    STATICS 


formed,  having  the  equilibrium   polygon  E'B'F'L'H'. 
The  line  of  action  of   the  resultant  R'  of  this  system 


FIG.  34 

intersects  the  line  of  action  of  the  resultant  R  at  the 
point  G,  which  point  is  the  centroid  of  the  system. 


CENTRE    OF    GRAVITY:   MOMENT   OF   INERTIA        115 

The  intersection  of  the  lines  of  action  of  resultants  is 
thus  the  general  method  of  finding  the  centre  of  gravity 
of  an  area  by  the  force  and  equilibrium  polygons,  the 
area,  if  possible,  being  first  divided  into  segments  of 
geometrical  form  whose  areas  and  the  locations  of  whose 
centres  of  gravity  are  known.  Then,  by  applying  at 
each  of  these  centres  of  gravity  a  force  proportional  to 
its  corresponding  area,  the  system  of  parallel  forces, 
shown  in  Fig.  34,  is  established. 

If  the  area  whose  centre  of  gravity  is  required  is  of 
irregular  shape  and  incapable  of  division  into  segments 
of  geometrical  form,  its  centre  of  gravity  may  be  located, 
in  close  approximation,  by  drawing  a  series  of  vertical 
lines,  closely  and  uniformly  spaced,  which  divide  the 
area  into  strips  so  narrow  that  the  centre  line  of  each 
space  may  be  regarded  as  the  length  of  that  strip.  At 
the  centre  of  gravity  of  each  strip,  i.e.,  the  middle  point 
of  its  centre  line,  apply  a  vertical  force  proportional  to 
the  area  of  the  strip,  that  is,  to  the  length  of  its  centre 
line,  since  the  strips  are  uniform  in  width.  Proceeding 
as  in  Fig.  34,  the  centroid  of  this  system  of  parallel  forces, 
which  is  the  centre  of  gravity  of  the  area,  can  be  found. 

26.  Moment  of  Inertia :  Radius  of  Gyration,  (a)  Mo- 
ment of  Inertia.  If  a  force  of  magnitude  P  act  at  a  per- 
pendicular distance  /  from  a  given  point,  its  moment 
about  that  point  is  P  x  /.  As  has  been  stated  previously, 
moments  can  be  treated  like  forces  in  graphic  processes. 
Hence,  if,  on  the  original  line  of  action,  the  moment 
P  X  /  be  assumed  to  act,  its  moment  about  that  point 
will  be  PI  x  /=  /Y2,  which  is  the  rectangular  moment 


Il6  GRAPHIC   STATICS 

of  inertia  /  of  the  force  P  about  that  point.  The 
moment  of  inertia  is,  therefore,  the  moment  of  the  first 
moment,  or  the  second  moment. 

Again,  if  a  particle  of  mass  m  be  rotating  with  an 
angular  velocity  v,  in  a  plane  about  a  point  lying  in  that 
plane  and  at  a  distance  r  from  the  particle,  the  angular 
momentum  of  the  particle  will  be  mvr^,  and  that  of  all 
the  particles  composing  the  body  will  be  v^mr2.  The 
expression  'Zmr2  =  /  is  the  sum  of  the  second  moments 
or  moments  of  inertia  of  the  particles,  m  being  the 
quotient  of  W,  the  weight,  divided  by^,  the  acceleration 
of  gravity.  Since,  in  the  mechanics  of  engineering, 
force  is  taken  as  the  product  of  mass  by  acceleration, 
this  term  may  be  used  to  describe  the  second  moment 
of  a  force  or  system  of  forces  about  an  axis ;  as  an  area, 
however,  is  not  a  material  body,  the  term  is  applicable 
strictly  to  it  only  on  the  assumption  that  each  ele- 
ment of  area  has  a  mass  proportional  to  its  area,  as  if 
the  given  figure  were  a  thin  plate. 

In  the  expression  for  /,  as  given  above,  the  radius  r 
evidently  differs  for  each  particle  considered.  Hence, 
the  moment  of  inertia  of  a  body  is  the  summation  of  the 
products  of  the  masses  of  the  elements  of  the  body  by 
the  squares  of  their  respective  distances  from  the  axis 
of  inertia  about  which  the  body  is  assumed  to  rotate. 
Similarly,  the  moment  of  inertia  of  a  force  about  such  an 
axis  is  the  product  of  the  square  of  the  distance  between 
the  point  of  application  of  the  force  and  the  axis,  by  the 
magnitude  of  the  force ;  and  the  moment  of  inertia  of  a 
system  of  parallel  forces  is  the  sum  of  these  products. 


CENTRE    OF    GRAVITY:    MOMENT    OF    INERTIA         Iiy 

The  moment  of  inertia  of  an  area  is,  in  the  same  way, 
the  summation  of  the  products  of  each  elementary  area, 
considered  as  a  mass,  by  the  square  of  its  distance  from 
the  axis  of  inertia. 

(b)  Radius  of  Gyration.  The  radius  of  gyration  is  the 
perpendicular  distance  from  the  axis  of  inertia  to  the 
centre  of  gyration.  For  a  body,  the  centre  of  gyration  is 
the  point  at  which,  if  the  entire  mass  of  the  body  were 
concentrated  in  a  single  particle,  the  effect  of  the  forces 
acting  on  the  body  would  be  unchanged  and  the  moment 
of  inertia  of  the  body  would  remain  the  same.  The 
centre  of  gyration  of  a  body  has  not  an  unchangeable 
location.  Its  position  may  be  taken  as  that  of  any  point 
in  the  body  which,  under  the  conditions  then  existing, 
is  at  a  distance  from  the  axis  equal  to  the  radius  of  gy- 
ration ;  with  any  variation  in  the  virtual  centre  about 
which  the  body  revolves,  that  radius  changes  also  and 
with  it  the  location  of  the  centre  of  gyration. 

If  we  assume  the  entire  mass  of  the  body  to  be  con- 
centrated at  the  centre  of  gyration,  at  a  distance  k  —  the 
radius  of  gyration  from  the  axis,  then  the  moment  of 
inertia  of  the  body  about  that  axis  is  : 


Hence  : 

JP  =  7/20*, 

i.e.,  the  square  of  the  radius  of  gyration  is  equal  to  the 
moment  of  inertia  of  the  body  divided  by  its  mass.  Simi- 
larly, the  square  of  the  radius  of  gyration  of  a  system  of 
parallel  forces  is  equal  to  the  moment  of  inertia  of  the 
system,  divided  by  the  magnitude  of  the  resultant  R  of 


n8 


GRAPHIC   STATICS 


the   system,  or  &  =  I/R.     For  an  area  A,  by  similar 
reasoning,   &  =  I/ A. 

27.  Moment  of  Inertia  of  a  System  of  Complanar  Par- 
allel Forces.  It  can  be  shown,  by  similar  triangles, 
that  the  moment  of  a  force  about  a  given  point  is  equal 
to  the  product  of  the  pole  distance  of  that  force  by  the 
intercept  which  is  cut  from  the  line  drawn  through  the 
given  point  and  parallel  to  the  line  of  action  of  the  force, 


FIG.  35 

when  the  two  sides  of  the  equilibrium  polygon  which 
intersect  on  the  line  of  action  of  the  force  are  prolonged. 
This  principle  furnishes  one  method  of  finding  the  mo- 
ment of  inertia  of  a  system  of  parallel  forces.  A  second, 
and  in  some  respects  more  convenient,  method,  that  by 
the  area  of  the  equilibrium  polygon,  is  given  below. 

In  Fig.  35,  let  Pl"-P3  be  a  system  of  complanar 
parallel  forces,  acting  from  the  points  of  application,  A, 
B,  and  C,  respectively.  Draw  the  force  polygon  a  ••-  dt 


CENTRE    OF    GRAVITY:   MOMENT   OF   INERTIA        IIQ 

with  pole  distance  H,  and  the  corresponding  equilibrium 
polygon  DEFG.  The  resultant  R  =  ad,  and  its  line  of 
action,  the  axis  Mm,  passes  through  the  point  G,  and 
hence  through  the  centroid  of  the  system.  For  any  of 
the  forces,  the  sides  of  the  polygon  which  intersect  on  its 
line  of  action  and  the  intercept  of  these  sides  on  the 
axis  form  a  triangle,  as  FGN  for  the  force  Ps,  to  which 
there  is  a  similar  triangle,  as  Ocd,  in  the  force  polygon. 
Let  /3  be  the  distance  between  the  line  of  action  of  /*8 
and  the  axis  Mm.  Then  : 

cd\GN  \\H\ly 
Since  cd  =  P  : 


Multiplying  by  /3  : 


P3/3  is  the  moment  of  inertia  /  for  the  force  Pz  about 
the  central  axis  Mm,  and  GNx  /3/2  is  the  area  of  the 
triangle  FGN.  By  similar  reasoning  : 

2  H  x  area  of  triangle  DGQ  =  I  for  Plt 
2  H  x  area  of  triangle  ENQ  =  /  for  P2. 
The  triangular  areas  : 

DGQ  +  ENQ  +  FGN=  polygonal  area  DEFG. 
Let  the  area  of  the  equilibrium  polygon  =  A'  .    Then: 

A1  X  2  H=  I  for  (P1  +  P2  +  PZ)  =/  for  R  ; 
and  for  the  system.     If  H  be  made  equal  to  R/2  : 

I=A'R, 
and  the  square  of  the  radius  of  gyration  is  : 

&  =  Af. 

In  Fig.  35,  the  forces  Pv  P2,  and  P3  have  all  the 
same  direction.  If  one  of  them,  as  P%,  had  a  direction 


120  GRAPHIC    STATICS 

opposite  to  that  of  the  other  two  forces,  its  moment  and 
the  corresponding  triangular  area  would  have  the  oppo- 
site sign,  and  the  area  A'  of  the  equilibrium  polygon 
would  then  be  equal  to  the  algebraic  sum  of  the  three 
triangular  areas. 

28.  Parallel  Axes  of  Inertia,  One  passing  through  the 
Centroid.  From  Art.  27,  it  will  be  seen  that  the 
moment  of  inertia  of  a  system  of  forces  about  an  axis 
passing  through  the  centroid  and  parallel  to  the  lines 
of  action  of  the  forces,  is  the  sum  of  the  moments  of 
inertia  about  that  axis  of  the  forces  between  whose  lines 
of  action  the  axis  lies.  This  principle  is  general,  not 
only  with  regard  to  an  area  and  a  body,  but  also  for 
any  given  axis  parallel  to  that  passing  through  the  cen- 
trode.  The  relation  between  the  moments  of  inertia 
about  two  such  parallel  axes  can  be  found  by  either  the 
method  of  intercepts  or  that  of  the  area  of  the  equilib- 
rium polygon.  The  latter  method  is  as  follows  : 

In  Fig.  35,  let  M'm'  be  any  given  axis,  parallel  to 
the  axis  Mm  through  the  centroid,  and  let  /j  be  the 
distance  of  the  line  of  action  of  the  force  Pl  from  the 
axis  M'm'  '.  The  triangles  Da'b'  and  Oab  are  similar. 
Hence  : 


a'b', 


=  moment  of  inertia  of  Pl  about  M'  m' 

=  r  for  force  /\. 

Hence,  /'  for  the  force  Pl  is  equal  to  2  H  x  area  of 
triangle  Da'b'.     Similarly  : 


CENTRE    OF    GRAVITY:   MOMENT   OF    INERTIA        121 

/'  for  force  P2  =  2  H  x  triangular  area  Eb'c' , 
I1  for  force  PB  =  2  H  x  triangular  area  /r'</. 
As  before,  let  the  area  of  the  equilibrium  polygon  =  A1 
and  that  of  the  triangle  Ga'd'  —  A".      Then,  the    mo- 
ment of  inertia  of  the  system  is : 

/'  =  2  H  X  polygonal  area  DEFGd'a1 

=  2ff(A'+A"). 

If  H  be  made  equal  to  one-half  the  resultant  R  of  the 
system  : 

I'  =R(A'  +  A"), 
and  the  square  of  the  radius  of  gyration  is : 

#*=A'  +  A". 

Hence,  for  an  axis  passing  through  the  centroid,  the 
moment  of  inertia  is  equal  to  the  product  of  the  re- 
sultant R  by  the  area  of  the  equilibrium  polygon,  if  the 
pole  distance  H  be  made  equal  in  magnitude  to  one- 
half  of  that  resultant.  For  a  parallel  axis,  the  magni- 
tude of  this  moment  is  increased  by  the  product  of  the 
resultant  by  the  triangular  area  formed  by  the  sides 
intersecting  on  the  line  of  action  of  the  resultant  and 
the  intercept  of  those  sides  on  the  given  axis.  For  a 
body  of  mass  M  or  a  cross-section  of  area  A,  the  re- 
sultant R,  as  above,  is  replaced  by  J/or  A,  respectively. 
29.  Moment  of  Inertia  of  an  Area.  —  While  the  gen- 
eral principles  established  in  Arts.  27  and  28  are  appli- 
cable fully  to  areas,  the  determination,  with  absolute 
accuracy,  of  the  moment  of  inertia  of  a  plane  area  by 
the  methods  there  given  is  impossible,  since  such  an 
area  consists  of  an  indefinitely  large  number  of  ele- 
ments of  area,  and  the  corresponding  system  of  parallel 


122  GRAPHIC   STATICS 

forces  would  therefore  be  composed  of  an  indefinitely 
large  number  of  forces,  thus  making  graphic  methods 
unavailable. 

(#)  Approximate  Determination.  —  A  working  approx- 
imation, which  will  serve  in  most  cases  in  practice,  may  be 
made  by  first  determining  the  centre  of  gravity  of  the 
given  area;  then  dividing  the  latter  into  a  number  of 
narrow  strips  parallel  to  a  complanar  axis  passing 
through  the  centre  of  gravity ;  and  finally  applying  at 
the  centre  of  gravity  of  each  strip  a  force  proportional 
to  its  area,  the  whole  forming  a  system  of  complanar, 
parallel  forces  which  can  be  treated  by  the  methods  of 
Art.  27.  The  greater  the  number  of  these  strips,  and 
hence  the  less  their  width,  the  nearer  the  approximation 
approaches  accuracy.  With  an  infinite  number  of  such 
forces,  it  is  evident  that  the  upper  sides  DEF  of  the 
equilibrium  polygon,  Fig.  35,  would  be  replaced  by  a 
curve  tangent  to  the  lower  sides  DG  and  FG.  In  prac- 
tice, if  the  given  area  be  divided  into  a  reasonably  large 
number  of  strips  as  explained,  this  curve  can  be  drawn 
with  sufficient  accuracy,  and  the  area  A1  of  the  equilib- 
rium polygon  can  then  be  measured  by  the  planimeter 
or  in  other  ways. 

(£)  Accurate  Determination.  Both  the  method  of  in- 
tercepts and  that  of  the  area  of  the  equilibrium  polygon, 
as  given  in  Arts.  27  and  28,  are  applicable  in  determin- 
ing, with  entire  accuracy,  the  moment  of  inertia  of  an 
area,  when  the  latter  can  be  divided  into  sections,  the 
area  of  each  of  which  and  its  moment  of  inertia  with 
respect  to  an  axis  passing  through  its  centre  of  gravity 


CENTRE   OF    GRAVITY:   MOMENT   OF    INERTIA        123 

are  known.  In  this  case,  the  force  representing  the 
area  of  the  section  is  applied,  not  at  the  centre  of  gravity 
of  the  latter,  but  at  a  distance  from  the  given  axis  of 
inertia  which  is  equal  to  the  radius  of  gyration  of  the 
area  of  the  section  about  that  axis. 

In  using  this  method,  a  further  relation  existing  be- 
tween the  moments  of  inertia  about  parallel  axes  of 
inertia  may  be  employed.  Thus,  let  7  be  the  moment  of 
inertia  of  a  system  of  parallel  forces  about  an  axis  Mm 
passing  through  the  centrode  of  the  system,  let  M'mf 
be  any  axis  parallel  to  Mm  and  at  a  distance  L  there- 
from, and  let  /'  be  the  moment  of  inertia  of  the  system 
about  the  new  axis  of  inertia  M*m'.  Then,  it  can  be 
shown  mathematically  that  : 

for  parallel  forces,  /'=/-}-  RL2, 

for  a  body,  I'  =f+ML*, 

for  an  area,  /'  =  7  +  AL2, 

R,  M,  and  A  being  the  resultant  of  the  forces,  the  total 
mass  of  the  body,  and  the  total  area,  respectively. 

Again,  let  a  be  the  area  of  any  one  of  the  sections 
into  which  the  total  area  is  divided,  k  its  radius  of  gyra- 
tion about  the  axis  Mm  passing  through  the  centre  of 
gravity,  k±  its  radius  of  gyration  about  the  given  axis  of 
inertia  M'm',  and  L  the  distance  between  the  two  axes. 
Then  (Art.  26)  : 


k*  =  fa2  +  L2  =  &2  +  L2. 
The  required  radius  of  gyration  is  therefore  the  hy- 


124  GRAPHIC    STATICS 

pothenuse  of  a  right-angled  triangle  whose  sides  are  k 
and  Z,  the  magnitudes  of  which  are  known.  The  force 
corresponding  with  the  sectional  area  a  is  then  assumed 
to  act  at  a  distance  k^  from  the  axis  of  inertia  and  to  be 
parallel  to  that  axis.  If  the  total  area  can  thus  be  di- 
vided into  geometrical  figures  to  which  this  principle 
can  be  applied,  either  the  method  of  intercepts  or  that 
of  the  area  of  the  equilibrium  polygon  can  be  used, 
with  entire  accuracy,  for  the  determination  of  the  mo- 
ment of  inertia  of  a  plane  area. 

EXAMPLES 

1 8.  CENTROID  OF  PARALLEL  FORCES  DUE  TO  LOCO- 
MOTIVE WHEEL  LOADS.  Let  it  be  required  to  find  the 
centroid  of  the  forces  due  to  the  locomotive  wheel  loads 
in  Example  15,  viz.  5.75  tons  on  the  pilot  wheel  and 
10.94  tons  on  each  of  the  four  drivers.  The  load-spac- 
ing, from  left  to  right,  is :  100,  66,  66,  and  72  inches. 

If  it  be  assumed  that  these  loads  are  held  in  equi- 
librium by  an  upward  force  P  equal  to  their  sum,  we 
shall  have  a  system  in  equilibrium  which  can  be  treated 
by  the  general  method.  In  Fig.  36  lay  off  the  five 
loads  on  the  span  segment  AA'  and  on  the  load  line  ab ; 
then  ba  —  P.  Draw  the  force  polygon  Oab  and  the  cor- 
responding equilibrium  polygon  BCDEFG.  The  verti- 
cal line  Mm,  passing  through  the  point  G,  is  then  the  line 
of  action  of  the  force  P  and  also  of  the  resultant  R  of 
the  forces,  which  resultant  is  equal  and  opposite  to  P 
and  acts  through  the  centroid  of  the  system.  This 
centroid  is  located  at  a  point  on  the  line  Mm  prolonged, 


CENTRE    OF    GRAVITY:    MOMENT    OF    INERTIA        1 25 


whose  location  depends  on  the  diameters  of  the  pilot 
and  driving  wheels.  Thus,  the  point  of  application  of 
the  resultant  of  the  forces 
on  the  four  driving 
wheels  lies  on  the  line 
of  centres  of  these 
wheels.  If  a  line  be 
drawn  from  this  point 
to  the  centre  of  the  pilot 
wheel,  the  intersection 
of  this  line  with  the  line 
Mm  prolonged  will  be 
the  centroid  of  the  sys- 
tem, or  the  point  of  ap- 
plication of  the  resultant 
R.  The  centroid  thus 
lies  about  10  inches  to 
the  right  of  the  centre 
of  the  second  driving 
wheel.  FIG.  36 

19.  CENTRE  OF  GRAVITY  OF  BULB  ANGLE.  When  a 
cross-section  has  an  axis  of  symmetry,  its  centre  of 
gravity  will  lie  on  that  axis,  and  but  one  equilibrium 
polygon  need  be  constructed.  The  bulb  angle  shown 
in  Fig.  37  is  not  thus  symmetrical,  and  hence  equilibrium 
polygons  must  be  drawn  for  two  directions  of  the  forces. 

The  dimensions  of  the  angle  are  :  total  depth,  6  inches; 
width  of  flange,  3  inches ;  thickness  of  web,  \  inch ;  of 
flange,  tapering  from  \  inch  to  ^g  inch ;  curves  of  bulb, 
\  inch  radius ;  of  flange,  ^g  inch  radius. 


126 


GRAPHIC   STATICS 


For  graphical  analysis,  the  cross-section  is  divided 
into  sections,  whose  areas  in  square  inches  are :  abc, 
0.12;  bcde,  1.02;  dhkl,  2.75;  efg,  0.02;  klo,  0.2;  pqr, 
0.05  ;  prso,  0.5  ;  rst,  0.4.  The  total  area  of  the  cross- 
section  is  thus  5.06  square  inches. 

At  the  centre  of  gravity  of  each  of  these  sections 


FIG.  37 

there  is  assumed  to  be  applied  a  force  equal  to  the  area 
of  that  section.  The  centre  of  gravity  of  a  rectangle 
lies  at  the  intersection  of  the  diagonals ;  of  a  triangle, 
on  a  line  drawn  from  any  vertex  to  the  middle  of  the 
opposite  side,  and  at  a  distance  from  the  vertex  equal  to 
two-thirds  the  length  of  this  line ;  of  a  circular  sector, 
on  its  middle  radius  at  a  distance  from  the  centre  of  the 
circle  equal  to  2  RC/$  A,  in  which  R  is  the  radius,  C  is 


CENTRE    OF   GRAVITY:   MOMENT   OF    INERTIA        127 

the  length  of  the  chord  of  the  arc,  and  A  is  the  length 
of  the  arc.  For  a  QO-degree  sector,  A  =  2  irR/^.  The 
centre  of  gravity  of  a  trapezoid  can  be  found  by  divid- 
ing it  into  two  triangles  and  connecting  their  centres  of 
gravity.  The  latter,  for  the  trapezoid,  will  lie  on  this 
connecting  line,  dividing  it  into  segments  inversely  pro- 
portional to  the  areas  of  the  two  triangles  so  that  the 
algebraic  sum  of  their  moments  about  it  shall  be  zero. 

In  Fig.  37,  the  forces  are  first  assumed  to  act  down- 
ward and  to  the  left.  The  centre  of  gravity  of  each 
section  is  located  and  a  force  equal  to  the  area  of  the 
section  is  laid  off  on  the  load  line  mn.  Assuming  equi- 
librium as  in  Example  18,  the  force  and  equilibrium 
polygons  Omn  and  ABCD  are  drawn.  The  resultant 
R  of  the  forces  acts  on  the  line  DG.  The  forces  are 
now  revolved  through  90  degrees,  and  the  force  and 
equilibrium  polygons  O'm'n'  and  A'B'C'D'  are  sim- 
ilarly constructed.  The  line  of  action  of  the  new 
resultant  R'  is  GD'.  The  two  lines  of  action  meet  at 
G,  which  is  the  centre  of  gravity  of  the  bulb  angle. 
The  point  G  is  vertically  distant  2.86  inches  from  the 
face  of  the  flange  and  3.14  inches  from  the  base  of  the 
bulb. 

20.  CENTRE  OF  GRAVITY  OF  A  PARTIAL  AREA.  The 
area  of  a  perforated  plate  is  its  total  area,  less  the  com- 
bined areas  of  the  perforations.  In  determining  the 
centre  of  gravity  of  such  a  plate,  the  area  of  each  per- 
foration must  evidently  be  treated  as  an  upward,  oppos- 
ing force,  in  order  to  consider  the  deductions  as  above. 

Thus,  Fig.   38,  let  it  be  required  to   determine  the 


128 


GRAPHIC   STATICS 


centre  of  gravity  of  the  partial  area,  ABCDEF,  which 
area  is  equal  to  that  of  the  rectangle  A  DBF,  less  those 

of  the  triangle 
ABC  and  the  circle 
g^K.  At  the  cen- 

tres  of  sravity«£"i> 

g<i,  gz  of  these  three 
figures,  apply  par- 


FIG. 38 


allel  forces  Plt  P>p 
and  P3  respective- 
ly proportional  to 
the  areas,  the 
forces  correspond- 
ing with  the  two 
deducted  areas 
being  assumed  to  act  in  the  opposite  direction  from 
Plt  which  represents  the  area  of  the  rectangle. 
Draw  the  force  polygon  and  the  equilibrium  polygons, 
LMNS  and  L'M'N'S',  for  two  directions  of  the  system 
of  forces.  The  resultants  R  and  R'  intersect  at  the 
point  G,  which  is  the  centroid  of  the  system  and  the 
centre  of  gravity  of  the  partial  area,  ABCDEF. 

21.  MOMENT  OF  INERTIA  OF  THE  CROSS-SECTION  OF 
A  DECK  BEAM  :  APPROXIMATE  METHOD.  —  Figure  39 
represents  the  cross-section  of  a  deck  beam.  It  is  re- 
quired to  determine  the  moment  of  inertia  of  this  cross- 
section,  about  an  axis  passing  through  the  centre  of 
gravity  of  the  cross-section  and  perpendicular  to  the 
web,  using  the  approximate  method  of  Art.  29. 

The  dimensions  are  :  depth  of  beam,  6  inches  ;  width 


CENTRE    OF    GRAVITY:   MOMENT   OF    INERTIA        I2Q 

of  flange,  4§  inches ;  thickness  of  web,  ^  inches ;  of 
flange,  |J  tapering  to  •£%  inches;  depth  of  bulb,  i^g 
inches;  thickness,  iJJ  inches;  area  of  cross-section,  4.1 
square  inches. 

For  the  reasons  stated  in  Art.  29,  the  cross-section  is 
divided   into    a    number   of   relatively   narrow   vertical 


FIG.  39 

strips,  and  the  area  and  centre  of  gravity  of  each  strip 
are  found.  It  is  assumed  that,  at  each  of  these  centres 
of  gravity,  there  is  applied  a  force  equal  to  the  area  of 
that  section,  the  whole  constituting  a  system  of  verti- 
cal parallel  forces,  whose  resultant  R  =  A  =4.1  square 
inches,  the  area  of  the  cross-section.  Assuming  equilib- 


130  GRAPHIC    STATICS 

rium  (Example  18),  the  force  and  equilibrium  poly- 
gons Omn  and  BCDE  are  drawn.  Since  the  cross- 
section  is  symmetrical  about  its  central  axis,  the  centre 
of  gravity  lies  on  this  axis,  the  line  of  action  of  the 
resultant  R  is  GE,  parallel  to  those  of  the  forces,  and  G 
is  the  centrode  of  the  system  of  forces  and  the  center  of 
gravity  of  the  cross-section. 

By  Art.  27,  the  moment  of  inertia  of  the  cross-section 
about  the  axis  GE  is  /=  RA1  —  AAf,  in  which  A1  is  the 
area  of  the  equilibrium  polygon.  It  is  evident  that, 
with  the  construction  shown  in  Fig.  39,  the  area  A',  as 
measured  from  the  drawing,  must  be  an  approximation 
only  to  the  true  area.  This  approximation  becomes 
closer  as  the  width  of  the  strips  is  decreased,  until, 
when  that  width  is  indefinitely  small,  the  broken  line 
BCD  becomes  a  curve  which  is  tangent  to  the  lines  EB 
and  ED,  prolonged,  at  the  points  b  and  d,  vertically 
below  the  ends  of  the  cross-section.  With  the  approxi- 
mate construction  as  in  Fig.  39,  the  measured  area  is 
thus  greater  than  the  true  area,  since  the  sides  of  the 
polygon  form  chords  of  the  true  curve.  In  this  case,  a 
full-scale  drawing,  made  with  fair  accuracy,  gave  an 
area,  measufed  by  planimeter,  of  ^'=5.34  square 
inches,  from  which  7=4.1  x  5.34  =  21.89,  as  against 
the  value  21.6  stated  in  the  manufacturer's  handbook 
and  obtained  from  an  area  bounded  by  the  true  curve. 

22.  ACCURATE  DETERMINATION  OF  THE  MOMENT 
OF  INERTIA  OF  AN  AREA  ABOUT  AN  Axis  PASSING 
THROUGH  ITS  CENTRE  OF  GRAVITY.  —  Let  it  be  required 
to  determine  accurately,  by  the  methods  of  Art.  29,  the 


CENTRE   OF    GRAVITY:   MOMENT   OF    INERTIA        131 


1 

i 

!  / 

X 

moment  of  inertia  of  the  geometrical  figure  shown  in 
Fig.  40,  about  an  axis  passing   through   its   centre  of 

gravity.     This  fig-        ^._ t ^r_ 

ure  is  composed  of     JT 
two  rectangles  and     /- 
a   triangle,    whose    ,| 
areas,  beginning  at  £  j~ 
the    left,    are :   a^     f. 
=  1.25,      rf2=i.5,     J^ 
a3=o.$6   square 
inches.     The  total 
area    is    therefore 
-4  =  3.31     square 
inches.       By  con- 
structing the  force 
polygon  Opq,  with 
pole  distance  A/2, 
and   the    corres- 
ponding   equilib-  FIG.  40 
rium    polygon  (not   shown),   the   centre    of  gravity  G 
of  the  figure  is  located. 

For  an  accurate  determination,  the  force  represent- 
ing an  area  should  be  applied  (Art.  29)  at  a  distance 
from  the  given  axis  of  inertia  equal  to  the  radius  of 
gyration  of  this  force  about  that  axis.  Further,  it  has 
been  shown  (Art.  29)  that,  if  kl  be  the  radius  of  gyra- 
tion of  an  area  about  an  axis  passing  through  its  centre 
of  gravity,  K^  the  similar  radius  about  a  parallel  axis, 
and  /j  the  distance  between  the  two  axes,  then 

1  —  *i  +  /i . 


132  GRAPHIC    STATICS 

In  this  case,  the  given  axis  of  inertia  is  M'm'  pass- 
ing through  the  point  G.  Applying  the  principles  as 
above  to  the  area  av  draw,  from  any  point  c  in  the 
axis  M'm1,  a  horizontal  line  intersecting  at  b  the  line  of 
action  gb  which  passes  through  the  centre  of  gravity 
of  alt  and,  from  b,  erect  bb'  =  kv  the  radius  of  gyration 
of  al  about  the  axis  gb.  Then,  cb  =  !1  and  cb'  —  V^-f/j2 
=  A"1,  the  radius  of  gyration  of  the  area  al  about  the 
axis  M'm' .  Revolving  cV  to  cb" ,  we  have  the  new  line 
of  action  b" B  of  the  force  corresponding  with  the  area 
alt  which  line  of  action  lies  to  the  left  of  the  original 
line. 

In  the  same  way,  the  new  lines  of  action  dD  and  eE, 
for  the  forces  corresponding  with  the  areas  #2  and  a3, 
respectively,  are  determined.  The  values  of  klt  £2,  and 
£3  are  found  analytically.  Thus,  the  radius  of  gyration 
of  a  rectangle  about  an  axis  passing  through  its  centre 
of  gravity  and  perpendicular  to  the  base,  is  equal  to 
the  length  of  the  base,  divided  by  the  square  root  of 
12;  the  similar  radius  for  a  triangle  about  an  axis 
parallel  to  the  base,  is  equal  to  the  altitude,  divided  by 
Vi8.  Hence,  k^  —  0.14,  >£2  =  0.87,  and  kz  =  o.i 8.  For 
clearness,  k^  and  k§  are  exaggerated  in  Figure  40. 

While  the  forces  have  thus  new  lines  of  action,  they 
have  still  the  same  magnitudes,  and  hence  the  original 
force  polygon  Opq  will  serve.  Constructing  the  corre- 
sponding equilibrium  polygon  BDEF,  the  line  of  action  of 
the  resultant  R  —  A  is  found  to  be  Mm  passing  through 
the  point  F.  The  problem  is  thus  reduced  to  the 
case  analyzed  in  Art.  28  and  Fig.  35,  viz.  there  are 


CENTRE    OF    GRAVITY:    MOMENT   OF    INERTIA        133 

two  parallel  axes  of  inertia  Mm  and  M'm',  the  former 
passing  through  the  centrode  of  the  system ;  the  mo- 
ment of  inertia  of  this  system  about  the  axis  M'm'  is 
therefore  /'  =  A(Ar  +  A"),  or  the  product  of  the  area 
of  the  cross-section  by  the  area  of  the  polygon  BDEF, 
plus  that  of  the  triangle  FJL,  formed  by  the  intersec- 
tion of  the  sides  BF  and  EF  and  the  intercept  cut  by 
those  sides  from  the  axis  M'm' .  Substituting  the 
value  of  the  area  A  and  those  of  the  two  latter  areas, 
as  measured  from  the  diagram,  I'  =  6.55. 


CHAPTER   VI 

FRICTION 

The  resistance  due  to  friction  is  both  helpful  and 
hurtful  in  the  performance  of  mechanical  work.  It  is 
friction  which  gives  the  driving  wheels  of  a  locomotive 
the  grip  on  the  rails  which  enables  them  to  move  the 
train ;  and,  again,  it  is  the  friction  of  the  car  wheels  on 
the  track  which,  disregarding  the  resistance  of  the  air, 
forms  the  total  work  of  pulling  the  train  and  develops 
the  total  stress  in  the  draw-bar.  The  analysis  of  the 
action  of  friction  is,  in  some  cases,  complex,  but  it  is 
often  simplified  materially  by  the  use  of  graphic 
methods. 

30.  Friction.  The  sliding  friction  of  solids  —  which 
only  will  be  treated  herein  —  is  the  resistance  to  relative 
motion  of  surfaces  in  contact  and  under  pressure.  This 
resistance  is  caused  by  the  interlocking  of  the  minute 
projections  and  indentations  of  these  surfaces.  If  the 
latter  were  absolutely  smooth  and  perfectly  hard,  there 
would  be  no  projections  to  disengage  and  override,  no 
frictional  resistance  would  occur,  and  hence  no  me- 
chanical work  would  be  required  to  produce  relative 
motion.  In  practice,  the  action  of  pure  friction,  as  above, 
is  complicated  by  adhesion,  abrasion,  the  viscosity  of 
lubricants,  etc. 


FRICTION  135 

Sliding  friction  is  the  friction  of  plane  surfaces  ;  the 
friction  of  warped  surfaces,  such  as  screw  threads,  and 
of  cylindrical  surfaces,  such  as  journals,  is  a  modified 
form  of  this  action.  Rolling  friction  is  the  friction  of 
a  curved  body,  as  a  cylinder  or  sphere,  when  moving 
over  a  plane  surface  or  one  of  greater  curvature.  In 
this  case,  contact  occurs  theoretically  on  a  line  or  point 
only  ;  but,  as  all  materials  are  more  or  less  elastic,  there 
is  actually  a  surface  of  contact,  and  therefore  rolling 
friction  is  identical  in  cause  with  sliding  friction.  The 
resistance  or  '  force  '  of  friction  acts  in  the  plane  of  con- 
tact of  plane  surfaces,  and,  with  curved  surfaces,  along 
their  common  tangent.  In  most  mechanisms,  the  fric- 
tion of  motion  only  is  encountered  ;  with  belting,  the 
friction  of  rest  occurs.  The  latter  is  greater  theoreti- 
cally than  the  former  since,  during  rest,  the  harder  body 
has  better  opportunity  to  indent  and  engage  the  softer 
surface  with  which  it  is  in  contact.  The  slightest  jar, 
however,  nullifies  this  action. 

The  force  of  sliding  friction  is  : 


in  which  F  is  the  total  resistance  or  force  which  opposes 
the  relative  motion  of  two  surfaces  in  contact,  -/Vis  the 
total  pressure  normal  to  those  surfaces,  and  /is  the 
factor  or  coefficient  of  friction.  This  formula,  which 
expresses  the  far  from  well-established  laws  of  sliding 
friction  —  assumes  that  the  total  force  of  friction  is  in- 
dependent of  the  area  of  the  surfaces  in  contact,  of 
their  relative  velocity,  and  of  the  intensity  of  pressure, 


136 


GRAPHIC    STATICS 


i.e.,  the  pressure  per  unit  of  surface.  According  to  this 
expression,  F  is  equal  simply  to  the  product  of  the  total 
normal  pressure  by  the  factor/  If  the  normal  pressure 
and  velocity  be  low,  and  the  surfaces  dry  or  but  slightly 
lubricated,  f  may  be  considered  as  having  a  constant 
value  for  the  same  materials  and  state  of  surfaces ;  but, 
in  the  wide  range  of  conditions  met  in  practice,  it  has 
been  found  that  the  value  of  f  is  affected  also  by  the 
velocity,  the  intensity  of  pressure,  the  temperature  of 
the  surfaces,  and  the  viscosity  of  the  lubricant,  so  that 
the  formula,  as  above,  is  incomplete  and  approximate. 
If,  however,  these  effects  be  all  considered  in  the  value 
assigned  to  the  factor  /,  and  this  value  be  determined 
independently  for  each  case,  the  formula  will  hold. 
From  the  formula  for  F,  we  have : 

/=  F/N=  tan  <£, 

i.e.,  the  coefficient  of  friction  is  the  tangent  of  an  angle 
known  as  the  angle  of  friction  c/>.  Thus,  in  Fig.  41,  let 
AB  and  AC  represent  planes  hinged  at  A  and  support- 
ing a  body  D  of  weight  W.  Keeping  AC  horizontal, 
let  B  be  raised  until  the  limiting 
condition  for  equilibrium  of  the 
body  D  is  attained,  that  is,  until 
with  any  further  elevation  of  B 
the  body  will  slide  downward. 
Let  the  angle  BAC=$-  Resolve 
^Finto  components  DE  and  EG, 
perpendicular  and  parallel,  respec- 
tively, to  the  surface  AB.  The  force  of  friction  F  is 


FRICTION  137 

produced  by  the  former  of  these  components  which  is 
the  total  normal  pressure  N  on  the  contact-surfaces,  and 
EG  is  equal  in  magnitude  and  opposite  in  direction 
to  F.  Hence : 

F=fW  cos  (j>  =  Wsin  </>,  and/=  tan  <f>. 

31.    Friction  of  Plane  Surfaces:    Friction  Cone.     Let 

Fig.  42  represent  a  body  of  weight  W  resting  on  a  hori- 
zontal    plane    AB.     While    no  K=N 
force   acts   but   that   of  gravity, 
the   body  will  be  held  in  equilib-    /»=/• 
rium   by   the  weight    W,  acting    A_ 
from  the  centre  of  gravity  O,  and 
by   the  reaction    R  =  N  of   the 
plane  AB.    This  reaction  is  equal  yr 
to  and  opposed  to    W,   has  the                  FlG'4* 
same  line  of  action,  is  normal  to  the   contact-surface, 
and  is  the  resultant  of  the  infinite  number  of  infinitely 
small  reactions  from  that  surface. 

Now  assume  a  force  P  =  Fas  applied.  Let  the  mag- 
nitude of  this  force  be  such  that,  with  any  increase,  it 
will  cause  motion  to  the  left,  the  limiting  condition  of 
equilibrium  having  been  reached ;  it  is  therefore  opposed 
by  the  full  force  of  friction  F.  The  body  is  now  in 
equilibrium  under  the  action  of  the  forces,  W,  R,  P,  and 
F.  The  resultant  of  R  and  F  is  the  virtual  reaction 
R'  or  CO,  making  the  angle  of  friction,  <£,  with  the  line 
of  action  of  R.  Since  equilibrium  prevails,  W,  R' ,  and 
P  meet  at  a  common  point  O.  For  motion  to  the  right, 
similar  but  reversed  conditions  exist,  as  is  shown  by 


138 


GRAPHIC    STATICS 


dotted  lines.  In  either  case,  the  supporting  plane  AB, 
the  normal  reaction  R,  and  the  force  of  friction  F  may 
be  replaced  by  the  virtual  reaction  R' ',  making  the  angle 
of  friction  with  the  normal  to  the  contact-surface,  and 
in  such  a  direction  that  its  component  parallel  to  that 
surface  will  oppose  the  relative  motion  of  the  two  bodies. 
It  will  be  seen  that  the  limiting  condition  for  equilib- 
rium, with  regard  to  motion  in  any  direction,  is  that  the 

virtual  reaction  shall  lie 
on  the  surface  of  the  cone 
described  by  the  revolu- 
tion of  the  triangle  OCD 
about  the  normal  OD. 
This  is  the  friction  cone 
or  cone  of  resistance. 
When  the  force  P  is  less 
than  F,  Rf  falls  within  the 
surface  of  the  cone  and 
there  will  be  no  motion ; 
when  P  is  greater  than  F, 
R1  lies  beyond  the  cone, 
and  motion  ensues. 

When  the  contact-sur- 
faces are  inclined,  the 
conditions  are  the  same  as 
with  level  surfaces,  ex- 
cept that,  owing  to  the 
FlG-43  inclination,  the  weight  W 

has  a  component  acting  along  the  incline  and  tending 
to  pull  the  body  downward.     Thus,  let  Fig.  43  represent 


FRICTION 


139 


an  inclined  plane  of  angle  BAC=Q,  on  which  a  body 
of  weight  W  rests.  In  the  upper  diagram  (A),  let  P 
be  a  force  acting  along  AB  and  just  sufficient  to  move 
the  body  upward  on  the  incline.  Momentary  equilibrium 
exists  under  the  action  of  the  weight  W,  the  normal  re- 
action R  which  is  equal  to  and  opposed  to  the  normal 
pressure  Nt  and  the  force  of  friction  F  which  acts  down- 
ward along  AB.  The  resultant  of  R  and  Fis  the  virtual 
reaction  R',  inclined  to  R  by  the  angle  of  friction.  Re- 
solving Wand  R',  parallel  and  perpendicular  to  AB,  we 

have : 

P  =  ac=ab  +  bc=  Wsin  0  +  F. 

In  diagram  (B\  let  Pf  be  the  force  which  is  just  suf- 
ficient to  move  the  body  downward  on  AB.  The  force 
F  now  acts  upward.  Resolving  as  before  : 

P' =  ac  =  ab-bc=W$m6-F. 

32.  Friction  of  Screw  Threads.  The  screw  thread  is 
essentially  but  an  inclined  plane  wrapped  around  a 
cylinder.  In  a  square  thread,  the  radial  elements  of 
the  thread  surface  are  perpendicular  to  the  axis  of  the 
cylinder;  in  a  triangular  thread,  these  elements  are 
inclined  to  that  axis.  The  axial  load  Wis  borne  usually 
by  the  bolt,  whose  thread  thus  corresponds  with  the 
contact-surface  of  the  body  D  in  Fig.  43 ;  the  nut 
thread  is  then  similar  in  its  action  to  the  surface  AB  in 
that  figure,  the  nut  being  supported  by  a  bearing 
surface. 

The  pressure  on  these  threads  is  assumed  to  be  con- 
centrated on  the  mean  helix,  or  the  circumference  of 


140 


GRAPHIC   STATICS 


the  mean  thread-diameter  d>  of  pitch  angle  a,  as  in 
Fig.  44.  Each  element  of  the  thread  surface  is  re- 
garded as  sustaining  an  equal  elementary  portion  of  the 
total  load  or  stress  W  on  the  bolt,  and  each  element 
has  therefore  a  frictional  resistance  of  the  same  magni- 


FlG.  44 

tude.  Since  the  conditions  for  all  elements  are  thus 
identical,  the  total  thread  resistance,  the  axial  load,  and 
the  external  turning  forces  on  the  nut  may  be  assumed 
to  be  each  equally  divided  and  concentrated  at  two 
points,  1 80  degrees  apart,  on  the  circumference  of 
diameter  d.  The  forces  P,  for  lifting  the  load,  thus 


FRICTION  141 

form  a  couple  whose  arm  is  d\  and  similarly  the  forces 
P't  for  lowering,  have  the  same  arm  and  points  of  ap- 
plication. In  Fig.  44,  these  points  are  //"and  K. 

(a)  Square  Threads.  In  Fig.  44,  taking  the  nut  as  the 
turning  member,  let  ABC  be  the  inclined  plane  formed 
by  developing  one  convolution  of  the  nut  thread  of 
mean  diameter  d\  AB  is  the  contact-surface  of  that 
thread,  and  EG  represents  a  portion  of  a  bolt  thread. 
The  base  of  the  plane  is  TT^,  its  height  is  the  pitch  /, 
and  the  pitch  angle  is  BAG.  Consider  the  forces  P  or 
P1  as  applied  to  the  nut  in  a  plane  normal  to  the  axis, 
and  as  tangent  to  the  mean  thread  circumference. 

When  the  nut  thread  is  on  the  point  of  moving  to  the 
left  to  raise  the  load,  its  front  half,  whose  center  of 
pressure  is  at  H,  is  in  equilibrium  under  the  action  of 
one  of  the  forces  P,  the  reaction  R'  making  the  friction 
angle  with  the  normal  reaction  R,  and  the  reaction 
Rl  =  W/2  of  the  nut  support,  which  reaction  is  vertical 
as  the  friction  of  the  support  is  neglected.  In  the  force 
polygon,  lay  off  ab  =  Rlt  and  draw  be,  bc\  and  ca  paral- 
lel, respectively,  to  the  lines  of  action  of  R't  R,  and  P. 

Then,  P  =  ca=  Wten  (0  +  a). 

When  the  nut  thread  is  about  to  lower  the  load  under 
the  action  of  the  force  P\  the  virtual  reaction  at  the 
contact-surfaces  becomes  R^ ',  making  the  friction  angle 
with  the  normal  reaction.  Equilibrium  then  exists  under 
the  action  of  the  force  Pf  and  the  reactions  Rl  and  RJ. 
Drawing  bd  parallel  to  the  latter, 

P  =  da  =  JFtan  (0  -  a). 


142  GRAPHIC    STATICS 

In  order  to  eliminate  all  friction  but  that  of  the  screw 
thread,  the  section  EG  of  the  bolt  thread  has  been 
assumed  to  have  no  lateral  motion. 

(b)  Triangular  Threads.  In  Fig.  44  a,  let  TV7"  and  N' 
be  the  normal  pressures  on  square  and  triangular  threads, 
respectively.  Then,  N'  =  TV' sec  ft,  in  which  ft  is  the 
base-angle  of  the  triangular  thread.  If  .Fand  F'  be  the 
forces  of  friction  for  the  two  threads,  we  have,  since 
F=fN: 

F'  =  fN'  =fN  sec  0  =  F  sec  0. 

Hence,  as  compared  with  the  square  thread  of  the 
same  pitch  angle,  the  friction  Ff  of  the  triangular  thread 
is  sec/3  times  greater.  To  determine  the  force  P  in 
Fig.  44  for  these  conditions,  prolong  the  lines  of  action 
R  and  R\  and,  at  any  point,  draw  the  line  ef  perpendic- 
ular to  R.  This  line  is  proportional  to  the  force  F. 
From  e  draw  eg  making  the  angle  @  with  ef\  from/ 
drop  the  perpendicular  fg  on  eg,  thus  determining  the 
line  eg  which  is  proportional  to  the  force  F1 '.  Revolve 
eg  to  eh  on  ef  prolonged,  and  draw  Hh  which  is  then  the 
line  of  action  of  the  virtual  reaction  R"  of  the  trian- 
gular thread.  In  the  force  polygon,  lay  off  bcn  parallel 
to  the  line  of  action  of  R",  and  the  force  P  will  then 
be  ac". 

33.  Pivot  and  Collar  Friction.  When  the  lower  end 
of  a  vertical  shaft,  subjected  to  end  thrust,  is  supported 
and  guided  by  a  step  bearing,  the  end  of  the  shaft  forms 
a  pivot  journal ;  the  latter  may  be  plane,  conical,  globu- 
lar, etc. 

(a)  Plane  Pivots.     This  form  is  shown  in  Fig.  45. 


FRICTION 


To  analyze  graphically,  let  the  total  load  on  the  shaft 
be  Wt  and,  as  with  the  screw  thread,  assume  it  to  be 
equally  divided  and 
one-half  concentra- 
ted at  each  of  the 
two  points  A  and 
Ar,  diametrically 
apart,  on  the  circle 
of  radius  r  des- 
cribed on  the  disk 
of  radius  rr  The 

total    normal    pres- 

i      i        •  FIG.  45 

sure  on  the  bearing 

is  W\  the  intensity  of  pressure,  or  average  pressure  per 
square  inch,  is  W I  irrf.  The  total  force  of  friction  is 
F  —  fW,  and,  assuming  uniform  pressure  over  the  disk, 
the  frictional  resistance  per  unit  of  area  isfW/Trr^. 

To  determine  the  relation  between  the  radii  ^  and  r, 
consider  the  circumference  2  trr  to  be  an  elementary 
ring  of  width  dr  and  of  area  2  irrdr.  The  total  normal 
load  on  this  ring  is : 

w  =  2  irrdr  x  W I  Trrf. 

Multiplying  w  by  the  coefficient  of  friction  f  gives 
the  frictional  resistance  of  this  elementary  ring,  and  this 
product,  multiplied  by  the  radius  rt  is  the  moment  of 
this  elementary  resistance,  or : 

»fr=m=*-fw. 


The  integral  of  this  expression,  between  the  limits 


144 


GRAPHIC    STATICS 


and  zero,  gives  the  moment  of  the  total  frictional  load 
or  force  Fy  which  is  : 


\      o  3 

Dividing  this  moment  by  its  force,  F  =  fW,  we  have 
the  mean  radius  at  which  F  acts,  or  : 

M  I  F  =  \r^  =  r. 

In  the  diagram,  BB  is  the  plane  of  the  disk,  CD  is 

one-half  W,  and  DE  and  DG  are  the  virtual  reactions 

at  the  points  A  and  A',  respectively. 

(b)  Conical  Pivots.     Fig.  46  represents  a  cone-shaped 

pivot.     Let  6  be  the  half-angle  of  the  cone,  r±  the  radius 

of  the  upper  end  of  the  jour- 
nal, and  Wthe  total  axial  load. 
As  with  the  plane  pivot,  con- 
sider one-half  of  W  as  con- 
centrated at  each  of  two 
points,  1  80  degrees  apart,  on 
an  elementary  strip  of  the 
conical  surface  of  radius  r  and 
width  dr.  Resolve  W  in  the 
directions  of  the  two  normals 
NN  to  the  contact-surface  at 

these  points.     Then,    N—  W  /2   sin  0,  and  the  total 

force  of  friction  is  : 


FIG.  46 


By  using  the  same  method  as  with  the  plane  pivot, 
the  mean  radius  at  which  the  total  force  of  friction  acts 
will  be  found  to  be  f  rr  Hence,  a  diagram  similar  to 


FRICTION 


H5 


that  in  Fig.  45  will  show  the  forces  graphically,  due 
regard  being  had  to  the  change  in  the  value  of  F. 

(c)  Mean  Radius  of  Friction.     For  the  two  cases  dis- 
cussed, this  radius  is  theoretically  f  rv  where  ^  is  the 
greatest   radius   of    the   bearing. 

This  is  true  of  all  pivot  bearings 
whose  projected  area  is  circular, 
and  not  annular.  This  theoreti- 
cal value  assumes  a  uniform  in- 
tensity of  pressure  and  a  constant 
value  of  f  for  the  whole  surface 
of  the  bearing.  These  conditions 
do  not  exist  in  practice,  and  a 
mean  radius  of  friction  equal  to 
one-half  the  greatest  radius  of  the 
bearing  surface  of  the  pivot  is 
generally  taken  as  more  correct. 
With  this  change,  the  formulae, 
as  deduced  above,  hold. 

(d)  Collar  Friction.     As  shown 

in  Fig.  47,  the  collar-bearing  is  simply  the  plane  pivot 
reduced  to  annular  form.  Let  W  be  the  total  axial  load 
on  the  shaft,  and  assume  it  to  be  equally  divided  and 
one-half  concentrated  at  each  of  the  points  A  and  A'. 
The  external  radius  of  the  bearing  is  r^\  its  internal  ra- 
dius is  rz ;  and  the  mean  radius  of  friction  is  r.  The 
value  of  r,  in  terms  of  r^  and  r2,  is  to  be  found. 

The  total  force  of  friction  is  fW,  and,  assuming  a 
uniform  pressure  over  the  whole  surface  of  the  bearing, 
the  frictional  resistance  per  unit  of  area  is 

L 


FIG.  47 


146  GRAPHIC    STATICS 

The  area  of  an  elementary  ring  of  radius  r  is  2  irrdr. 
The  total  load  on  this  ring  is : 

w  =  2  Trwtfr  X  W/7r(r?  —  r£\ 

The  frictional  resistance  of  this  load  is  fzu  and  the 
moment  of  this  resistance  is  : 

m  —fw  x  r=frx  2  irrdr  x  W/7r(r?  —  r£). 
The  integral  of  this  expression  between  the  limits  r^ 
and  r2  gives  the  moment  of  the  total  frictional  load  or 
force  F,  which  moment  is : 


Dividing  this  moment  by  the  force  F  —  fW,  the  mean 
radius  of  friction  is  : 

r    =  ?.*!- 

3    TI 

34.  Journal  Friction;  Friction  Circle.  —  (a)  Journal 
friction  is  complex  in  analysis,  since  the  fit  is,  in  most 
cases,  more  or  less  free,  and  pure  sliding  friction  be- 
tween cylindrical  surfaces  rarely  occurs  in  ordinary 
machinery,  although  in  the  mechanism  of  precision  it 
is  the  rule.  The  usual  shaft  journal  rests  in  its  bearing 
somewhat  loosely,  giving  theoretically  a  line  bearing  be- 
tween cylindrical  surfaces  of  different  radii,  which  is 
rolling  friction.  When  the  shaft  revolves,  friction 
causes  the  journal  to  roll  up  the  side  of  the  bearing, 
like  a  car-wheel  moving  up  an  inclined  track.  The 
journal  ascends  until  the  weight  makes  it  slide  back- 
ward, and  this  rise  and  fall  continue  with  every  varia- 
tion in  the  coefficient  of  friction.  The  method  of 
analysis  which  follows  is  that  first  established  by 


FRICTION 


147 


(a) 


Rankine  and  later  developed  by  Hermann  and  others. 
It  applies  to  close-fitting  bearings  only. 

In  Fig.  48,  let  A  be  a  cylindrical  journal  at  rest  in  the 
bearing  B\  the 
forces  acting  are 
the  weight  or 
other  vertical  pres- 
sure W=  CD  and 
an  equal  and  op- 
posite reaction 
from  the  bearing 
at  the  point  G. 
Now,  let  there  be 
applied  a  turning 
force  P,  normal  to 
a  radius  and  of 
such  magnitude 
that  the  limiting 
condition  of  equi- 
librium is  reached, 
the  frictional  re- 
sistance is  overcome,  and  the  journal  is  on  the  point  of 
beginning  clockwise  rotation.  The  forces  acting  are 
then  W,  P,  and  the  virtual  reaction  R'  from  the  bearing, 
whose  line  of  action  is  yet  to  be  determined. 

Prolong  the  lines  of  action  of  W  and  P  until  they 
meet  at  C;  lay  off  CD  =  W  and  DE  equal  to  P  and 
parallel  to  the  latter's  line  of  action.  Then,  CE,  the 
resultant  of  Wand  P,  is  equal  in  magnitude,  opposed  in 
direction,  and  has  the  same  line  of  action  as  the  virtual 


148  GRAPHIC    STATICS 

reaction  R'  from  the  bearing,  since  IV,  P,  and  R'  are 
in  equilibrium  and  their  lines  of  action  must  meet  at  a 
common  point.  Draw  CEH,  and,  from  the  centre  of 
pressure  K,  lay  off  KH  —  R1 .  This  reaction  is  simply 
the  resultant  of  the  indefinite  number  of  indefinitely 
small  reactions  from  the  bearing. 

Draw  the  line  of  action  OL  of  the  normal  reaction  R 
through  the  point  K\  resolve  R'  normally  and  tangen- 
tially.  Then,  R  =  LK  is  the  total  normal  pressure  on 
the  journal,  and  the  force  of  friction  is  F=  HL  —  KL  x 
coefficient  of  friction  /.  Since  the  virtual  and  normal 
reactions  are  inclined  from  each  other  by  the  angle  of 
friction,  the  angle  HKL  =  OKE  =  <£. 

(b)  Friction  Circle.  In  Fig.  48,  drop  the  perpendicu- 
lar OM  from  the  centre  O  on  the  line  of  action  of  the 
reaction  R' .  Then,  if  r  be  the  radius  of  the  journal 
OM '  —  r  sin  (f>.  If  now  the  radial  arm  6W  and  the  force 
P  be  revolved  through  all  positions  about  the  centre  O, 
it  will  be  found  that,  while  the  location  of  the  point  of 
intersection  C  will  be  changed,  the  perpendicular  distance 
of  the  line  of  action  of  R'  from  the  centre  will  be  equal 
in  every  case  to  r  sin  <£.  Hence,  the  locus  of  such 
points  of  intersection  as  M  is  a  circle  described  from  O 
and  of  radius  rsin  <£.  This  is  \.\IQ  friction  circle. 

The  principle  is  general,  applying  to  all  forms  of 
loading  of  a  closely  fitting  journal  and  bearing,  either 
of  which  has  motion  in  any  direction  relatively  to  the 
other.  The  dotted  lines  show  the  effect  of  anti-clock- 
wise rotation,  the  force  P  having  the  new  line  of  action 
CN1 9  but  the  other  conditions  remaining  the  same.  If 


FRICTION  149 

it  be  assumed  that  the  journal  is  fixed  and  the  bearing 
revolves  —  like  a  connecting  rod  pulling  a  crank  pin  — 
the  magnitude  and  lines  of  action  of  R'  will  be  the  same 
for  rotation  in  either  direction,  but  the  centres  of  pres- 
sure will  be  changed  from  K  and  Q  to  K'  and  Q't 
respectively.  Again,  assume  that  the  pressure  acts  up- 
ward instead  of  downward,  as  at  (a),  Fig.  48,  and  that 
the  rotation  is  clockwise,  the  force  P  acting  from  W 
to  C.  If  the  bearing  be  the  moving  member,  the  centre 
of  pressure  will  be  at  Q ,  at  which  point  R'  will  act 
downward  on  the  journal;  if  the  journal  be  the  moving 
member,  the  load  will  act  upward  and  R'  downward  at 
the  point  Q.  It  will  be  seen  that  the  friction  circle  and 
a  tangent  to  it,  which  is  the  line  of  action  of  the  virtual 
reaction  R',  wholly  replace  either  the  journal  or  the 
bearing  in  the  graphical  investigation  of  their  friction. 
The  general  laws  of  sliding  friction  may  be  applied  to 
determine  on  which  of  these  tangents  R'  acts  in  any 
given  case. 

35.  Link  Connections:  Friction  Axis.  A  link  is  a 
straight,  rigid  machine  member  employed  to  transmit 
power  between  two  rotating  or  oscillating  members,  or 
one  rotating  and  one  sliding  member,  as  is  the  case 
with  the  connecting  rod.  Journal  friction  therefore 
exists  at  each  end  of  the  link ;  the  principles  governing 
its  action  are  those  established  in  Art.  34.  If  there 
were  no  friction,  the  radius  of  the  friction  circle  would 
be  zero  and  the  force  would  be  transmitted  from  one 
member  to  the  other  along  the  line  joining  the  centres 
of  the  journals  or  bearings  of  the  link.  With  friction, 


GRAPHIC   STATICS 


the  resultant  of  the  transmitted  force  and  the  force  of 
friction  acts  along   the  friction  axis,  which  is   a   line 

tangent  to  the  friction  circles 
of  the  two  journals. 

There  are  four  fundamen- 
tal cases  of  this  action,  two 
of  which  are  represented  in 
Figs.  49  and  50  and  the  others 
deduced  therefrom.  In  each 
of  these  mechanisms,  A  is  a 
link  and  B  and  C  are  hinged 
levers,  P  is  a  force  acting  up- 
ward on  the  lower  or  driving 
lever,  and  the  dotted  lines 
show  the  path  in  ascending. 
It  will  be  observed  that : 

(a)  The  link  may  be  assumed  to  be  in  momentary 
equilibrium  in 
any  given  posi- 
tion, but  its  effi- 
ciency, as  thus 
determined,  ap- 
plies to  that  posi- 
tion only.  Also, 
the  direction 
of  the  friction 
axis  depends  on 
which  of  the  con- 
nected levers  is 
the  driver,  and 


"vx 

/        \\ 

(0) 

_f__    J 

Y 

p 

Fig.  50 

FRICTION  151 

hence  on  whether  the  link  is  in    tension   or   compres- 
sion. 

(b)  A  force  and  its  reaction  may  be  regarded  as  two 
equal   and   opposite   forces    having   the    same   line   of 
action.      Since   the  link  is  in  equilibrium,  the  friction 
axis  is  the  line  of   action  of   a  force  acting  from  one 
journal  and  an  opposing  reaction  from  the  other. 

(c)  The   link   bearings  move  on  the   journals  which 
they  support,    the   journals   being   thus   relatively  sta- 
tionary.    The  direction  of  this  motion  is  determined  by 
that  of  the  mechanism. 

(d)  Since  the  link  is  in  equilibrium,  the  force  of  fric- 
tion at   either   journal  must   so  act   as   to  oppose   the 
relative   motion   of    the   bearing.      This    consideration 
determines   the   direction  of   the  friction  axis.     If  the 
direction  of  rotation  is  the  same  with  both  bearings,  the 
axis  will  lie  diagonally,  being   tangent  to  one  friction 
circle  on  one  side,  and  to  the  other  on  the  opposite  side. 
If  the  bearings  rotate  in  opposite  directions,  the  friction 
axis  will  be  parallel  to  the  geometric  axis. 

In  Fig.  49,  the  lower  lever  is  the  driver.  Hence, 
disregarding  friction,  the  force  on  the  lower  journal  is 
concentrated  at  the  point  a  on  the  geometric  axis,  act- 
ing there  on  the  lower  bearing  of  the  link.  This  puts 
the  latter  in  compression,  and  its  force  acts  on  the 
lower  side  of  the  upper  journal  at  the  corresponding 
point  b.  From  the  path  of  the  mechanism,  the  motion 
of  both  bearings  is  seen  to  be  anti-clockwise.  At  the 
lower  bearing,  the  tangential  force  of  friction,  F=de, 
acts  from  left  to  right,  and  therefore  on  the  left  side 


152  GRAPHIC    STATICS 

of  the  friction  circle,  where  it  and  the  normal  pressure 
line  cd  combine  to  form  the  resultant  pressure  line  ce. 
Hence,  the  friction  axis  is  here  tangent  to  the  left  side 
of  the  friction  circle. 

At  the  upper  journal,  F  must  act  from  right  to  left, 
combining  with  the  reaction  from  the  journal  to  form 
the  virtual  reaction  which,  acting  along  the  line  ec, 
must  be  tangent  to  the  right  side  of  the  friction  circle. 
If,  on  the  other  hand,  the  upper  lever  be  the  driver  in 
lifting  the  connected  parts,  the  direction  of  motion  of 
the  bearings  will  be  the  same,  but  the  link  will  be  in 
tension  and  the  friction  axis  will  be  tangent  to  the  other 
side  of  each  friction  circle. 

In  Fig.  50,  the  lower  lever  is  the  driver  and  the  link 
is  again  in  compression,  but  the  direction  of  rotation  of 
the  two  bearings  is  not  the  same.  Hence,  the  forces  of 
friction  act  on  the  same  sides  of  the  two  friction  circles 
and  the  friction  axis  is  parallel  to  the  geometric  axis. 
If,  again,  the  upper  lever  be  the  driver  in  lifting  the 
connected  parts,  the  direction  of  motion  of  the  bearings 
will  not  be  changed,  but  the  link  will  be  in  tension,  and 
the  friction  axis,  while  still  parallel  to  the  geometric 
axis,  will  be  tangent  to  the  other  sides  of  the  friction 
circles. 

36.  Chain  Friction :  Resistance  of  Ropes  to  Bending. 
(a)  Chains.  The  friction  of  a  chain  in  passing  over  a 
chain  drum  or  sprocket  wheel,  is  a  modified  form  of 
journal  friction,  the  pins  joining  the  links  constituting 
the  journals.  Let  Fig.  51  represent  a  chain  pulley,  of 
effective  radius  R,  revolving  in  a  clockwise  direction. 


FRICTION 


The  load  on  the  advancing  or  left  side  of  the  chain  is 
W',  the  driving  force  on  the  right  or  receding  side  is  P. 
Let  rbe  the  radius 
of  the  pins  joining 
the  links,  and  r^  that 
of  the  wheel  journal. 

Relative      motion      Bl 
of  the  links   of  the    £(\ 
chain     occurs    only 
when    the    latter 
bends  at   the  joints 
on   reaching  and 
leaving  the  horizon- 
tal  diameter  EF  of  Flr"  5I 

the  wheel.  At  this  time,  the  advancing  link  A  turns  in 
an  anti-clockwise  direction  on  the  link  B,  the  latter  serv- 
ing as  a  bearing;  and,  similarly,  the  link  C  is  rotated  in 
the  same  direction  with  regard  to  the  link  D.  From 
Art.  34,  it  will  be  seen  that  the  effect  of  these  actions  is 
to  remove  the  lines  of  action  of  the  load  W  and  the 
force  P  from  the  vertical  lines  passing  through  the 
centres  of  the  link-pins  over  to  the  left  by  the  distance 
rsin<f>t  i.e.,  the  friction  of  the  joints  of  the  chain  in- 
creases the  leverage  of  the  load  and  decreases  that  of 
the  driving  force  by  that  amount.  Similarly,  the  line 
of  action  of  R',  the  vertical  reaction  of  the  wheel-bear- 
ing, is  removed  to  the  right  through  the  distance  r  sin  <£. 
R1  is  the  resultant  of  P  and  W.  Hence  : 


R  —  (r-f  r^sin  <f) 


154  GRAPHIC   STATICS 

(b)  Ropes.  The  resistance  of  a  rope  in  passing  on  or 
off  a  sheave  or  grooved  pulley  has  an  effect  similar  to 
that  of  chain  friction  on  the  lines  of  action  of  the  load 
and  driving  force,  i.e.,  the  lever  arm  of  the  load  is  in- 
creased, and  that  of  the  driving  force  decreased,  by  the 
same  amount  in  both  cases.  This  effect  is  not  due, 
however,  to  the  same  causes  as  with  the  chain.  In  the 
rope,  it  is  produced,  when  the  latter  is  bent,  by  the  rela- 
tive motion  and  consequent  friction  of  the  strands,  by 
the  compression  of  the  fibres  on  the  inner  side  and  the 
stretching  of  those  on  the  outer  in  winding  on,  and  by  the 
reversal  of  this  action  when  the  rope  leaves  the  pulley. 

Thus,  if  we  assume  the  rope  to  be  wound  on  a  pulley, 
like  the  chain  in  Fig.  51,  it  will  be  bent  in  advancing 
when  it  passes  at  E  above  the  horizontal  diameter  EF, 
and  will  be  straightened  again  when  it  descends  below 
F  at  the  right.  In  the  first  of  these  operations,  the 
normal  forces  acting  on  the  horizontal  cross-section  of 
the  rope  at  E  are  the  tension  at  the  centre  due  to 'the 
load,  the  tension  on  the  outer  half  from  the  stretching 
of  those  fibres,  and  the  opposing  compressive  force  on 
the  inner  half.  The  resultant  of  these  three  forces  is  a 
tension  which  acts  at  a  distance,  which  for  convenience 
we  shall  call  s,  outward  from  the  centre  of  the  rope,  so 
that,  neglecting  the  friction  of  the  pulley  journal,  the 
lever  arm  of  the  weight  W  is  R  4-  s,  R  being  the  dis- 
tance from  the  centre  of  the  pulley  to  that  of  the  rope. 

When  the  rope  straightens  at  Ft  the  same  forces  act, 
but  those  of  tension  and  compression  from  bending 
change  places,  occurring,  respectively  on  the  inner  and 


FRICTION  155 

outer  halves  of  the  cross-section.  Hence,  the  resultant 
tensile  force  acts  between  the  pulley-  and  rope-centres, 
at  the  distance  s  from  the  latter,  so  that,  neglecting 
journal  friction,  the  lever  arm  of  the  driving  force  P  is 
R  —  s.  It  will  be  seen  that  the  distance  s  thus  corre- 
sponds with  the  radius  of  the  friction  circle  for  chain 
friction  ;  and,  as  with  the  latter  and  considering  journal 
friction,  we  may  write  : 

R  +  r  sin     +  s 


R  —  r±  sin  (f>  —  s 

in  which  r±  sin  </>  is  the  radius  of  the  friction  circle  for 
the  journal  of  the  pulley. 

The  value  of  s  can  be  expressed  only  by  empirical 
formulae,  the  constants  of  which  have  been  derived 
from  experiment.  This  follows,  since  the  resistance  of 
a  rope  to  bending  varies  directly  as  the  tensile  stress  in 
it  due  to  the  load  or  the  driving  force  ;  directly  as  some 
power  of  its  diameter,  since  the  smaller  the  rope,  the 
greater  its  pliability  ;  and  inversely  as  the  radius  of  the 
pulley,  since  the  greater  this  radius,  the  less  the  re- 
quired bending.  Evidently,  it  also  depends  upon  the 
material  of  the  rope,  its  length  of  service,  etc. 

37.  Belt  Gearing.  The  friction  of  a  belt  on  a  pulley 
in  power  transmission  is  the  friction  of  rest  and  not 
that  of  motion  ;  it  is  not  the  cause  of  lost  work,  but  the 
means  by  which  useful  work  is  done  in  preventing  rela- 
tive motion  of  the  working  parts;  and  the  greater 
within  practical  limits  it  becomes,  the  better.  With 
regard  to  the  friction  of  motion  of  belt  gearing  —  that 
of  the  pulley  shaft  in  its  bearing  —  the  reverse  is  true, 


156 


GRAPHIC    STATICS 


its  efficiency  being  much  less  than  that  of  toothed  gear- 
ing, since  the  power  transmitted  is  directly  proportional 

to  the  difference  between 
the  tensions  of  the  tight 
and  slack  sides  of  the  belt, 
while  the  aggregate  thrust 
on  the  bearing  which  pro- 
duces journal  friction  is 
due  to  the  pull  of  both  of 
these  tensions. 

The  graphics  of  belt 
gearing  present  no  prob- 
lems which  have  not  been 
examined  previously  here- 
in. Figure  52  represents 
a  pulley  A  driving  through 
belting  a  pulley  B,  the  ro- 
tation being  anti-clockwise. 
Each  pulley  may  be  con- 
sidered as  in  equilibrium 
under  the  action  of  a  driv- 
ing force  or  a  resistance, 
the  two  belt  tensions,  the  normal  reaction  of  the  bearing, 
and  the  journal  friction,  the  latter  being  relatively  large. 
The  angle  of  contact  0  between  belt  and  pulley  can  be 
found  by  construction,  and  the  ratio  between  the  two  ten- 
sions can  be  determined  from  the  well-known  formula : 

7\=£/9' 
in  which  the  angle  0  is  given  in  circular  measure,  E  is 


FIG.  52 


FRICTION  157 

the  base  of  the  Napierian  system  of  logarithms,  T2  and 
7\  are,  respectively,  the  tensions  of  the  driving  and 
slack  sides  of  the  belt  during  motion,  and  /  is  the 
coefficient  of  friction. 

Prolong  the  lines  of  action  of  these  tensions  until 
they  intersect  at  O ;  and,  from  O,  lay  off  OC  and  OD 
of  such  dimensions  that  OC/OD  is  equal  to  the  tension 
ratio  just  found.  Their  resultant  OE  gives  the  line  of 
action  of  the  resultant  T  of  T2  and  Tlt  which  line  in- 
tersects the  lines  of  action  of  the  resistance  W,  of  lever- 
arm  /,  and  the  driving  force  Pt  of  arm  Z,  at  the  points 
F  and  G,  respectively.  From  the  latter  points  and 
tangent  to  the  friction  circles,  draw  the  lines  of  action 
of  RI  and  Re,,,  the  virtual  reactions  of  the  bearings. 

The  driven  pulley  B  is  in  equilibrium  under  the  ac- 
tion of  the  resistance  W,  the  resultant  T  of  the  tensions, 
and  the  virtual  reaction  Rlt  the  magnitude  of  W  and 
the  lines  of  action  of  all  the  forces  being  known.  In 
the  lowest  force  triangle,  lay  off  be  =  W  and  draw  ba 
and  ca  parallel,  respectively,  to  the  lines  of  action  of 
T  and  Rlt  thus  determining  the  magnitudes  of  the  two 
latter.  To  find  the  magnitude  of  P,  consider  the  driv- 
ing pulley  as  similarly  in  equilibrium  and  lay  off,  in  the 
second  triangle,  ab  =  T  and  draw  ad  and  ab  parallel, 
respectively,  to  the  lines  of  action  of  P  and  R%.  The 
magnitude  of  the  tensions  is  determined  in  the  upper 
force  triangle  by  resolving  the  resultant  T  of  the  ten- 
sions parallel  to  the  lines  of  action  of  its  components,  7\ 
and  TV  The  effect  of  journal  friction  can  be  shown  by 
replacing  Rl  and  Rz  by  reactions  passing  through  the 


158 


GRAPHIC    STATICS 


centres  of  the  respective  journals,  and  drawing  the  corre- 
sponding sides  of  the  force  triangles  parallel  thereto, 
which  method  will  give  the  values,  excluding  journal 
friction,  of  P,  T,  Tv  and  T2  for  the  same  resistance  W. 
38.   Friction  of  Gear  Teeth.  —  In  the  transmission  of 
power  by  gear  wheels,  a  part  of  the  lost  work  is  ex- 
pended   in    overcoming 
the  frictional  resistance 
of    the  teeth.      This  is 
due  to  the  fact  that  a 
pair  of  teeth,  while  en- 
gaged, move   one  upon 
the    other,   the    line   of 
bearing    changing   con- 
tinually on  both  contact 
surfaces  and  the  relative 
motion     of     the     latter 
being  a  combination  of 
rolling  and  sliding. 

Figure  53  (a)  repre- 
sents the  positions,  at 
the  beginning  and  end 
of  engagement,  of  the 
same  pair  of  involute 
teeth  on  the  spur  gears  A  and  B,  the  wheel  A  being 
the  driver  and  rotating  in  a  clockwise  direction.  The 
two  pitch  circles  are  tangent  at  the  point/  on  the  line 
joining  the  centres  of  the  wheels.  The  arcs  bp  and  cp 
are  the  arcs  of  approach ;  ptf  and  pc1  are  the  arcs  of 
recess.  The  line  aa*  is  the  line  of  action,  which  is  the 


FIG.  53 


FRICTION  159 

path  of  the  points  of  contact  of  the  teeth  during  engage- 
ment. This  line  is  normal  to  the  contact-surfaces  and 
in  properly  formed  teeth  it  always  passes  through  the 
point/,  making  an  angle  with  the  horizontal  called  the 
angle  of  obliquity  of  action. 

In  approach,  the  points  b  and  c  gradually  draw  nearer 
until  they  meet  at/.  Since  the  arc  ab  is  shorter  than  the 
arc  ac,  it  follows  that  the  flank  of  the  driving  tooth  rolls 
through  a  distance  equal  to  ab  on  the  face  of  the  driven 
tooth,  and  slides  for  a  distance  equal  to  ac  —  ab.  In  recess ', 
this  process  is  reversed,  the  face  of  the  driver  rolling  on 
the  flank  of  the  driven  tooth  through  a  distance  equal  to 
c'a* ,  and  sliding  through  the  distance  b' a1  — c'a' . 

Figure  53  (b)  shows  the  engaged  portions  of  the  two 
gear  wheels  A  and  B.  In  this  figure,  it  is  assumed  that 
the  lengths  of  the  arcs  of  action,  bpb1  and  cpc\  in  (#),  are 
such  that  two  pairs  of  teeth  are  simultaneously  engaged, 
and  that  the  normal  pressure  is  the  same  between  the 
teeth  of  each  pair.  The  line  of  action  is  aa'  as  before. 

If  there  were  no  friction,  the  reaction  due  to  the  load 
on  the  driven  teeth  3  and  4  would  act  on  the  drivers  i 
and  2  along  the  line  a1  a  from  right  to  left,  and  this  re- 
action would  be  exactly  equal  to  the  force  P,  exerted 
by  the  driving  wheel  and  acting  in  the  direction  aa'  and 
on  that  line.  From  the  enlarged  force  polygon  (c),  it 
will  be  seen  that,  considering  friction,  the  reaction  from 
tooth  3  takes  the  direction  da,  and  that  from  tooth  4, 
the  direction  a' d,  both  reactions  being  inclined  to  the 
normal  by  the  angle  of  friction  <£  and  intersecting  the 
line  of  centres  at  the  point  d.  If  these  virtual  reactions 


160  GRAPHIC    STATICS 

be  resolved  vertically  and  parallel  to  aa' ',  the  vertical 
components  will  neutralize  each  other  and  the  true  line 
of  action  of  the  force  and  the  resultant  reactions  will 
become  a^a^  parallel  to  afa  and  at  a  perpendicular  dis- 
tance therefrom  of  tan  $  x  aaf '/2,  which,  assuming  aa'  to 
be  equal  to  the  circular  pitch,  becomes  tan  <£  x  pitch/2. 

The  friction  of  spur  gear  teeth  is  then,  in  its  effect, 
similar  to  chain  friction.  With  frictionless  motion,  both 
the  force  and  resistance  would  act  along  the  normal  aa1 ; 
with  friction,  the  line  of  action  is  shifted  to  the  parallel 
line  a^a\.  This  change  increases  the  distance  from  the 
centre  of  gear  A,  at  which  the  load  from  gear  B  acts  on 
A,  and  decreases  the  distance  from  the  centre  of  gear  B, 
at  which  the  driving  force  from  gear  A  acts  on  B. 
Hence,  in  general,  the  leverage  of  the  power  is  lessened 
and  that  of  the  load  is  increased  by  the  distance  deter- 
mined above,  tan  </>  x  pitch/2. 

Figure  53  and  this  discussion  refer  to  involute  teeth. 
With  cycloidal  teeth,  the  line  of  action  is  not  a  straight 
line,  as  aa',  but  is  an  arc  of  the  circle  with  which  the 
profiles  of  the  teeth  are  described.  The  loss  from  tooth 
friction  and  the  wear  in  service  are,  in  consequence,  less 
than  with  any  other  form  of  tooth. 

EXAMPLES 

23.  FRICTION  OF  STATIONARY  ENGINE.  The  readiest 
practical  approximation  to  the  power  expended  in  over- 
coming the  friction  of  an  engine  can  be  obtained  by 
taking  indicator  cards  with  the  engine  unloaded  and 
running  at  various  speeds.  There  are  two  important 
objections  to  this  method :  the  errors  of  the  indicator 


FRICTION 


161 


and  the  fact  that  the  results  found  include  all  resistances, 
so  that  it  is  impossible  to  separate  that  for  any  particu- 
lar bearing.  On  the  other  hand,  the  difficulty  with 
analytical  or  graphic  methods  is  that  the  coefficient  of 
friction  cannot  be  estimated  with  much  accuracy,  since 


FIG.  54 

it  appears  to  vary  not  only  with  pressure,  but  with  veloc- 
ity, the  temperature  and  condition  of  the  contact-sur- 
faces, the  viscosity  of  the  lubricant,  etc.  The  graphic 
method  for  the  crosshead,  crank  pin,  and  shaft  bearings 
is  shown  in  Fig.  54. 

The  data  are:  cylinder  diameter,  15  inches;  stroke, 
1 8  inches  ;  length  of  connecting  rod,  36  inches ;  maximum 
unbalanced  pressure  on  piston,  100  pounds  per  square 
inch ;  corresponding  total  force  acting  on  piston  before 
cutoff,  17,671  pounds. 


1 6  2  GRAPHIC    STATICS 

In  the  position  shown  in  Fig.  54,  the  crosshead  is 
approaching  the  mid-point  of  the  forward  stroke,  and 
hence  the  angle  with  the  horizontal  made  by  the  con- 
necting rod  is  increasing,  while  that  between  the  connect- 
ing rod  and  crank  is  decreasing.  Only  centre  lines  and 
the  friction  circles  are  shown.  In  computing  the  radius 
of  these  circles,  the  same  diameter,  6  inches,  will  be 
taken  for  the  crosshead,  crank  pin,  and  shaft,  since  the 
friction  circle  is  so  small  that  even  a  considerable  varia- 
tion in  these  diameters  will  affect  but  little  its  radius, 
in  the  graphic  method  illustrated  here.  The  crosshead 
slipper  reciprocates  and  the  bearing  of  the  crosshead 
pin  oscillates,  while  the  crank  pin  and  shaft  are  rotating 
journals.  The  coefficient  of  friction  will  hence  vary 
considerably  for  each.  For  convenience,  an  average 
coefficient, /=  0.185,  will  be  taken  for  all,  with  a  conse- 
quent angle  of  friction  </>  of  10°  30'.  Then,  the  radius  r 
of  each  journal  is  6/2  =  3  inches,  and  the  radius  of  the 
friction  circle  is  r  sin  <£  =  3  x  0.182  =  0.546  inch. 

From  the  principles  given  in  Art.  35,  the  small 
arrows,  showing  the  direction  of  the  frictional  resistance, 
are  described  at  each  bearing,  and  the  friction  axis  AB 
is  drawn,  as  the  line  of  action  of  the  thrust  T  on  the 
connecting  rod.  The  crosshead  pin  is  in  equilibrium 
under  the  action  of  Tt  the  piston-force  P,  and  the  virtual 
reaction  R^t  inclined  to  the  left  by  the  angle  <£  from  the 
normal  Rl  to  the  slipper  bearing.  These  three  forces 
must  meet  at  the  common  point  A,  which  is  the  inter- 
section of  the  lines  of  action  of  P  and  T;  this  determines 
the  location  of  RJ.  In  the  force  polygon,  lay  off  ab  =  P 


FRICTION  163 

and  draw  ac  and  be,  parallel  respectively  to  the  lines  of 
action  of  R\  and  T.     Then,  be  =  T. 

The  direction  of  the  line  of  action  of  the  resistance 
of  the  load  to  the  motion  of  the  shaft  depends  upon 
what  the  latter  drives  and  upon  where  the  power  is 
transmitted  from  this  driven  member.  For  compact- 
ness in  the  drawing,  assume  in  this  case  that  there  is 
a  gear  wheel,  2  feet  in  diameter,  fixed  to  the  shaft  and 
engaging  a  similar  wheel  which  is  set  on  the  engine 
centre-line  and  between  the  shaft  and  crosshead.  It 
is  required  to  determine,  for  the  shaft-driven  gear,  the 
magnitude  of  the  resistance  W,  of  arm  L  =  I  foot, 
which,  at  this  position  of  the  engine,  the  force  P  will 
overcome. 

The  crank,  the  shaft,  and  the  gear  wheel  fixed  to 
the  latter  form  virtually  a  bell-crank  whose  horizontal 
arm  is  of  length  L.  This  bell-crank  is  acted  on  by  the 
forces  T  and  W,  the  normal  reaction  R2  from  the  bear- 
ing, and  the  frictional  resistance  to  the  shaft's  motion, 
which  resistance  acts  in  a  clockwise  direction.  The 
lines  of  action  of  T  and  W  meet  at  the  point  C  and  the 
virtual  reaction  R<£  from  the  shaft  bearing  acts,  by 
Art.  35,  on  a  line  drawn  from  C  and  tangent  to  the 
upper  side  of  the  friction  circle.  In  the  force  polygon, 
the  intersection  of  cd  and  bd,  drawn  parallel  to  the  lines 
of  action  of  R\  and  W,  respectively,  give  W-=bd  — 
11,000  pounds. 

Disregarding  friction,  the  forces  are  T ,  Rlt  and  R^t 
as  is  shown  by  dotted  lines.  Constructing  the  force 
polygon  for  these  conditions,  we  have  be—  W=  13,200 


1 64 


GRAPHIC    STATICS 


G   H  K 


F'         ^~r< 

— i\  //v1 

w-^l^J-r^  F 


pounds.     For  this  position  of  the  engine,  its  efficiency 
is   therefore    11/13.2  =  83.3   per   cent,  and   the   power 

expended  in  over- 
coming the  friction 
of  these  four  bear- 
ings is  16.7  per 
cent  of  the  total 
force  P  on  the  pis- 
ton. 

24.  FRICTION  OF 
SCREW  JACK.  Fig- 
ure  55  represents 
the  upper  part  of 
an  ordinary  screw 
FIG'55  jack.  The  screw 

A  engages  the  nut  B,  fixed  in  the  supporting  frame 
C\  the  load  W  rests  on  the  swivel  plate  D,  journalled 
on  the  screw  head  E,  the  latter  being  bored  for  the  bar  by 
which  power  is  applied  to  turn  the  screw  and  lift  the  load. 
The  data  are :  nominal  or  outside  diameter  of  screw, 
2^  inches ;  depth  of  thread,  J  inch ;  diameter  at  base 
of  thread,  2  ^g  inches ;  pitch  of  screw,  J  inch ;  pitch 
angle  «  of  mean  helix  (Art.  32),  4°,  about. 

Carefully  conducted  laboratory  experiments  on  a  well 
oiled  screw  jack  gave,  for  the  resistance  of  the  thread,  a 
coefficient  of  friction  /=  0.1494  =  tan  <£,  and  hence  an 
angle  of  friction  <f>  of  8°3o'.  These  results  are  doubt- 
less somewhat  lower  than  those  usual  in  practice. 

Let  the  circle  of  radius  r±=  i-f%  inches  be  the  hori- 
zontal projection  of  the  mean  helix  and  assume  the  load 


FRICTION 


W  to  be  equally  divided  and  its  halves  concentrated  at 
the  upper  and  lower  extremities  of  the  vertical  diameter 
of  this  circle.  If  F  be  the  horizontal  component  of 
the  thread  resistance  on  one-half  of  the  mean  helix  and 
P  be  the  total  horizontal  force  acting,  equally  divided 
as  a  couple,  at  the  ends  of  the  diameter  2  r^  on  the  bar 
to  raise  the  load, 
then  one-half  or 
P/2  will  be  taken 
as  acting  at  the 
same  points  as 
F. 

The  screw  ro- 
tates in  an  anti- 
clockwise direc- 
tion to  raise  the 
load,  the  bolt  thread  thus  moving  upward  on  the  sta- 
tionary nut  thread.  The  conditions  are  then  as  shown 
in  Fig.  56,  in  which  BAC\s>  the  pitch  angle,  AB  is  the 
upper  surface  of  the  nut  thread,  and  D  is  a  portion 
of  the  bolt  thread  representing  one-half  of  a  convolu- 
tion. This  section  D  is  in  momentary  equilibrium 
under  the  action  of  the  load  W/2t  the  virtual  reaction 
R'  making  the  angle  of  friction  <j>  with  the  normal 
reaction  R,  and  the  horizontal  raising  force  P/2.  Hence  : 

</>), 


=-  =  ~  tan  (a 


which  is  also  the  magnitude  of  the  force  F=  GK  (Fig. 
55),  acting  at  each  of  the  two  points  of  application  of 
the  load  Won  the  mean  helix. 


1 66  GRAPHIC   STATICS 

The  tangent  of  a  +  </>  is  0.222.  Taking  W  as  10  tons 
(20,000  pounds)  P/2  =  2220  pounds,  which  is  one  of  the 
forces  of  the  horizontal  couple  acting  at  the  extremities 
of  the  diameter  2  rx  =  2-fg  inches  to  raise  the  load ; 
it  is  also  equal  to  the  force  F  of  the  resisting  couple 
acting  with  the  same  arm. 

This  calculation  assumes  that  the  load  rotates  with 
the  jack  screw,  which  is  seldom  the  case.  If  the  load 
be  stationary,  the  friction  of  the  swivel  plate  D  must  be 
considered.  The  mean  radius  of  friction  (Art.  33)  of 
this  plate  is  (Fig.  50)  r2  =  i  \  inches.  Taking  the  same 
angle  of  friction,  and  assuming  that  one-half  of  the  load 
W  is  concentrated  at  each  of  two  points  diametrically 
apart,  as  before,  the  force  of  friction  F'  (GH'm  Fig.  50) 
acting  at  the  extremities  of  the  diameter  2  r^  is  10,000  x 
0.1494  =  1494  pounds. 

There  are  thus  two  couples  which,  for  simplicity,  may 
be  considered  as  acting  in  the  same  plane:  P/2  =  F— 
22 20  pounds  with  arm  of  2^  inches,  and  F*  =  1494 
pounds  with  arm  of  2|  inches.  These  two  couples  may 
be  assumed  to  act  at  any  points  in  their  respective  cir- 
cumferential paths.  Taking  the  points  of  application 
as  90  degrees  apart,  as  in  Fig.  55,  the  resultant  couple 
is  found  to  have  a  force  F"  =  2666  pounds  with  an  arm 
d  of  3.32  inches.  If  the  raising  bar  be  of  such  length 
that  its  couple,  as  actually  applied,  has  an  arm  of  24 
inches,  the  force  required  at  each  of  its  ends  will  be 
about  368  pounds. 

Since  it  is  not  possible  to  make  the  load  W  absolutely 
central,  there  will  also  be  journal  friction  on  the  pin  L 


FRICTION 


I67 


on  which  the  swivel-plate  is  centred.  The  amount  of 
this  resistance  will  depend  on  the  eccentricity  of  the 
load  and  the  resulting  normal  pressure  on  the  pin.  If 
this  pressure  be  known,  the  frictional  resistance  can  be 
determined  by  the  methods  of  Art  34. 

25.  PULLEY  BLOCKS  :  RELATION  OF  LOAD  AND 
POWER.  Professor  Hermann1  gives  the  method 
shown  in  Fig.  57  for  finding  the  relation  of  load  and 
power,  and  the  ten- 
sions in  the  vari- 
ous portions  of  a 
chain  or  rope  pass- 
ing  over  the 
sheaves  of  the  pair 
of  blocks  of  a 
tackle,  friction  be- 
ing considered. 

In  Fig.  57,  the 
blocks  are  three- 
sheaved  ;  the  load 
W  is  suspended 
from  the  lower  or 
movable  block  A, 
and  the  frame  C  of 
the  upper  or  fixed 


FIG.  57 


block  B  is  supported  from  above.  The  rotation  of  the 
sheaves  in  raising  the  weight  is  clockwise,  the  rope 
winding  on  at  D  and  E  and  off  at  F  and  G.  If  the 

1 "  Graphical  Statics   of    Mechanism,"   Hermann-Smith,   New   York, 
1904,  p.  88. 


1  68  GRAPHIC    STATICS 

upper  or  fixed  end  be  secured  to  the  hanger  C,  as  at 
Tv  there  will  then  be  seven  portions  of  the  rope  whose 
tensions,  7\  to  T7,  are  to  be  determined  —  the  last,  or 
that  of  the  hauling  end,  being  equal  to  the  lifting  force 
P. 

For  any  two  consecutive  sections  of  these  seven  — 
one  on  each  side  of  a  sheave  of  either  block  —  the  ten- 
sion in  the  rope  on  one  side  constitutes  the  load,  and 
that  on  the  other,  the  power  for  the  pair.  If  R  be  the 
effective  radius  of  the  sheave,  r^  that  of  the  block- 
journal,  and  /  be  equal  to  r  sin  <£  or  to  s,  as  in  Art.  36, 
then  the  relation  between  the  power  /  and  the  load  w 
on  these  two  portions  will  be,  for  either  a  chain  or  a 
rope: 


R  -  ^  sin  <t>  -  t 

At  the  journal  of  the  lower  block,  the  load  W  acts 
on  the  vertical  tangent  to  the  left  (Art.  35)  of  the 
friction  circle  ;  at  the  upper  journal,  the  reaction  of  the 
support  C  acts  on  the  similar  tangent  to  the  right. 
Again,  from  the  figure,  it  will  be  seen  that  the  unwind- 
ing or  power  side  is  the  left  on  the  lower  block  and  the 
right  on  the  upper,  the  tensile  forces  in  the  sections  of 
chain  or  rope  acting,  in  each  case,  at  a  distance  R  —  t 
from  the  centre,  as  shown  by  the  full  lines  ab  and  cd\ 
similarly,  on  the  respectively  opposite,  or  load,  sides  of 
the  blocks,  the  tensile  forces  act  at  a  distance  R  +  t 
from  the  centre.  Hence,  as  the  lower  block  is  free,  it  will, 
when  ascending,  swing  to  the  left  for  a  distance  2  t,  as 
shown  in  the  figure,  so  that  the  tensile  forces  shall  act 


FRICTION  l6p 

vertically.  When  upward  motion  ceases,  the  block 
swings  backward  until  it  is  again  vertically  below  the 
upper  block  ;  when  the  load  is  lowered,  the  process  is 
reversed. 

In  passing  from  the  fixed  end  of  the  chain  or  rope 
to  the  free  or  hauling  end,  its  tension  is  increased, 
whenever  it  passes  over  a  sheave,  in  the  ratio  given  by 
the  reciprocal  of  the  fraction  in  the  preceding  equation. 
Hence,  starting  at  the  fixed  end,  the  general  expression 
for  the  ratio  between  the  tensions  in  two  consecutive 
sections  is : 

Tn   _  R  -  r^  sin  <ft  -  /  (, 

Tn+l      R  +  ^  sin  $  +  /' 

In  the  left-hand  figure,  draw  the  horizontal  line  HK, 
cutting  the  lines  of  action  of  the  rope-tensions  ab  and 
cd  at  K  and  H,  respectively,  and  those  of  the  load  W 
and  the  reaction  from  the  upper  bearing  at  <?2  and  olt 
respectively.  Then,  HK  =  2  R  and  the  distances  : 

Hol  =  Ko<i  =  R  —  r^  sin  <j>  —  t, 
0^2=  2  r^  sin  <f>  -f  2  /*, 
Ho<i  =  Hol  4-  o  0*1  —  Ko^  =  R  -f  r±  sin  (/>  +  /. 

Assume  that  the  tension  7^  in  the  first  section  from 
the  fixed  end  is  known  and  is  equal  to  H-i  on  the  line 
cd.  From  j,  draw  through  ol  the  line  1-2  cutting  the 
line  ab  at  2.  Then,  K-2  is  equal  to  the  tension  T2  in 
the  next  succeeding  section,  for  the  triangles  H,  i,  o^ 
and  Ky  2,  ol  are  similar,  and  : 

ff,  I  __  ffol  __  R  —  r^  sin  (ft  —  t  _  T^  ,  , 

K,  2  ""  Ko  ~~  R  +  T  sin  <   -f-  /""  T 


1 70  GRAPHIC    STATICS 

Continuing  in  the  right-hand  diagram,  draw  from  the 
point  2  through  02  the  line  1-2  and  by  similar  reasoning 
H-3  is  the  tension  Tz  in  the  third  section.  In  the  same 
way,  K-4  =  F4,  H-5  =  T6,  K-6  =  TQ,  and  H-7  =  T7  =  P 
are  found.  The  sum  of  the  tensions,  7\  to  TQ,  inclusive, 
which  is  the  line  LM,  is  equal  to  the  load  W,  while  the 
force  P  is,  on  the  same  scale,  equal  to  MN. 

Since  the  distances  Holt  Ho^,  Ko^  and  Ko^  are  con- 
stant for  any  system  of  sheaves  such  as  is  shown  in  the 
figure,  it  is  evident  that  the  ratio  MN/LM  is  constant, 
and  that  any  value  of  7\  may  be  assumed  in  finding  the 
ratio.  When  the  latter  is  determined,  the  value  of  P 
for  any  given  value  of  Wis  given  by  the  expression : 

P^W- MN/LM. 

As  an  example,  consider  a  3|-inch  hemp  rope  (ij 
inches  diameter),  passing  over  a  pair  of  3-sheaved 
blocks,  as  in  Fig.  57.  Take  the  power  P,  applied  at 
the  hauling  end,  as  1000  pounds,  or  slightly  above 
800  pounds  per  square  inch  of  sectional  area  of  the 
rope ;  and  let  the  effective  radius  R  of  the  pulley,  i.e., 
that  to  the  centre  of  the  rope,  be  6  inches,  the  diameter 
of  the  pin  or  shaft  of  each  block  be  i^  inches  making 
r^  =  |  inch,  and  the  coefficient/  of  journal  friction  be 
o.i,  giving  (f>  —  6°,  about.  It  is  required  to  determine 
the  weight  W  which  the  given  power  P  will  lift. 

It  is  necessary  first  to  find  the  value  of  /,  in  order  to 
draw  the  lines  of  action  ab  and  cd.  This  value  is  also 
that  of  s  in  Art.  36.  It  cannot  be  calculated  with  any 
great  degree  of  accuracy  for  reasons  given  previously. 


FRICTION  171 

Various  empirical  expressions  have  been  used.  Eytel- 
wein's  formula,  as  employed  by  Reuleaux  and  Weisbach,1 
gives,  for  the  total  stiffness  or  resistance  to  bending  5 
of  a  hemp  rope  in  both  winding  on,  and  unwinding 
from,  a  pulley : 

5  =  0.472  Wd*/R, 

in  which  W  is  the  load  and  d  and  R  are  respectively 
the  diameter  of  the  rope  and  the  radius  of  the  pulley, 
both  in  inches. 

Assuming  that  the  resistance  to  winding  is  equal  to 
that  of  unwinding  —  which  is  only  approximately  true 
—  we  have,  for  either  operation  : 

5/2  =  0.236  W<P/R. 

Now,  if  the  rope  were  absolutely  elastic,  5  would  not 
exist,  and  the  moment  of  the  weight  would  be  W  X  R. 
Considering  stiffness,  this  moment  is,  by  Art.  36, 
W(R  +  s).  The  added  moment  Ws  is  therefore  equiva- 
lent to  that  produced  by  S/2  acting  at  the  radius  R. 

Hence:  Ws  =  SR/2. 

Substituting  : 

s  =  0.236  d?  inch. 

While  the  results  given  by  this  formula  are  somewhat 
high,  it  is  sufficiently  accurate  for  large  hemp  rope 
under  heavy  stress.  For  rope  of  other  materials,  the 
constants  require  modification.  Substituting  the  value 

of  d,  i^  inches  : 

s  =  /  =  0.368  inch. 

1  "  Mechanics  of  Engineering  and  Machinery,"  New  York,  1896,  Vol.  I, 

§  197- 


172  GRAPHIC    STATICS 

With  <t>  =  6°,  the  radius  of  the  friction  circle : 
rl  sin  <£  =  0.625  x  0.1045  =  0.065  inch. 

In  Fig.  57,  draw  the  upper  sheave,  the  friction  circle 
and  the  dotted  tangent  at  G.  Let  fall  the  tangent  at  B 
until  it  meets  at  o1  the  indefinite  horizontal  line  HK. 
Draw  the  line  of  action  cd  parallel,  to  the  left,  and  at 
a  distance  /  from  the  tangent  at  G\  this  line  meets  HK 
at  Ht  thus  locating  the  latter  point.  From  ol  set  off  to 
the  left  the  distance  o^  =  2(^  sin  </>  +  /)  =  0.866  inch ; 
from  <?2,  drop  the  tangent  A  to  the  lower  friction  circle, 
and  describe  the  latter  to  its  right,  thus  determining 
the  centre  of  the  lower  sheave.  Erect  the  dotted  tan- 
gent at  the  left  of  this  sheave,  and,  at  a  distance  /  to 
the  right  of  it,  draw  the  line  of  action  ab,  intersecting 
HK&K. 

Assume  any  value — say  500  pounds  —  for  H-i  and 
lay  it  off  on  the  line  Hd.  Then,  as  explained  pre- 
viously, draw  1-2,  2-3,  3-4,  etc.  Adding  H-i,  K-2, 
ff-3,  K~4,  H~5>  and  K-6,  and  dividing  their  sum  by 
H-7,  we  have  3.7,  which  is  the  value  of  the  ratio  W/P. 
Hence  W—  3700  pounds.  If  stiffness  and  friction 
were  not  considered,  the  value  of  this  ratio  would  be  6. 
The  efficiency  of  the  tackle  is  thus  61.66  per  cent,  which 
is  a  fairly  accurate  result.  This  method  assumes  that 
the  section  7\  of  the  rope  is  vertical  and  thus  parallel 
to  the  other  sections. 

The  value  of  the  ratio  in  Equation  (3)  is  0.865. 
This  value  is  constant  throughout  the  system,  and  gives 
the  relation  between  the  tensions  in  any  two  consecutive 
sections  of  the  rope. 


FRICTION 


173 


26.   SPUR  GEARS  :  RELATION  OF  LOAD  AND  POWER. 

Figure   58  represents  the  pitch  circles  of  a  train  of 
spur  gears,  A,  B,  and  C,  with  involute  teeth.     The  power 
applied  to  the  driving  gear  A  is  equivalent  to  a  force  P 
of  arm  L  ;  the  resistance  acting 
on  the  gear  C  is  equal  to  a  force 
Wot  arm  L1.     Assuming  fric- 
tion at  the  wheel  journals  and 
between  the  engaged  teeth,  it 
is    required    to    determine   the 
magnitude  of  the  force  P  for 
a  known  resistance  W. 

The  driving  gear  A  is  in 
equilibrium  under  the  action 
of  the  force  P,  the  reaction 
7*!  from  the  teeth  of  gear  B, 
and  the  virtual  reaction  A\  from 
the  bearing.  By  Art.  38,  the 
reaction  7\  is  parallel  to  the 
line  aa'  passing  through  the 
point  of  tangency  of  the  pitch 
circles,  the  distance  of  7\  from 
the  centre  of  A  being  greater 
than  that  of  aa1  by  the  amount 
given  in  the  article  cited.  The 
lines  of  action  of  P  and  7^  intersect  at  c,  and  the  reac- 
tion Rl  drawn  from  c  is,  by  Arts.  34  and  35,  tangent  to 
the  lower  side  of  the  friction  circle. 

The  intermediate  gear  B  is  in  equilibrium  under  the 


FIG.  58. 


174  GRAPHIC    STATICS 

action  of  the  force  Tv  the  reaction  T2  from  the  teeth  of 
gear  C,  and  the  virtual  reaction  R^  from  the  bearing. 
Since  the  gear  B  is  the  driver  for  gear  C  and  its  motion 
is  opposite  to  that  of  gear  A,  the  reaction  T2  has,  by 
Art.  38,  the  direction  and  location  shown  in  the  figure, 
being  parallel  to  the  line  bb'  passing  through  the  point 
of  tangency  of  the  pitch  circles.  The  lines  of  action  of 
TI  and  T2  intersect  at  dt  from  which  point  the  reaction 
R2  is  drawn  tangent  to  the  lower  side  of  the  friction 
circle. 

The  driven  gear  C  is  in  equilibrium  under  the  action 
of  the  resistance  J/F,  the  force  7"2,  and  the  reaction  R% 
from  the  bearing.  The  lines  of  action  of  T2  and  W 
meet  at  ^,  from  which  point  the  reaction  Rz  is  drawn 
tangent  to  the  upper  side  of  the  friction  circle. 

The  directions  of  all  the  forces  acting  on  the  train  and 
the  magnitude  of  one,  the  resistance  W,  are  known. 
Starting  with  W,  the  force  triangle  for  each  gear  and 
the  force  polygon  for  the  train  can  be  drawn  by  the 
general  method,  as  shown  by  the  diagram  below. 

For  these  gears,  the  angle  of  obliquity  (Art.  38)  is 
usually  about  15  degrees,  that  is,  the  line  bb'  makes  an 
angle  of  15  degrees  with  the  horizontal. 


PROBLEMS 

1.  What  weight  can  be  drawn  up  an  inclined  plane  rising  I  in  5 
by  a  pull  of  200  pounds :   (a)  when  the  pull  is  parallel  with  the 
plane ;   (b}  when  it  is  horizontal  ?     Disregard  friction. 

2.  A  wheel  of  5  feet  diameter  weighs  2  tons,  including  its  load. 
What  is  the  least  horizontal  force  necessary  to  pull  it  over  a  stone 
4  inches  high  ? 

3.  A  boiler  weighing  5000  pounds  is  supported  by  tackles  from 
the  fore  and  main  yards.     If  the  tackles  make  angles  of  30  and  35 
degrees,  respectively,  with  the  vertical,  what  is  the  tension  in  each 
of  the  two  ? 

4.  Sheer  legs  50  feet  long  are  spread  18  feet  at  the  base.     The 
back  stay  is  80  feet  long.     Find  the  stresses  acting  in  each  member 
when  lifting  a  load  of  25  tons  at  a  distance  of  20  feet  from  the  foot 
of  the  sheer  legs,  the  weight  of  the  latter  being  disregarded. 

5.  A  crane  post  is  12  feet  high  ;  its  jib,  32  feet  long,  and  stay,  23 
feet  long,  meet  at  the  peak  A.     Two  backstays,  making  angles  of  45 
degrees  with  the  horizontal,  are  in  planes  due  north  and  due  west 
from  the  post.     A  weight  of  6  tons  is  suspended  from  A.     Find  the 
forces  in  the  jib  and  stays :   (a)  when  A  is  southeast  of  the  post ; 
(b)  when  A  is  due  east;  (<:)  when  A  is  due  south. 

6.  A  rod  weighing  10  pounds  rests  in  a  smooth  hemispherical 
bowl  which  is  fixed  with  its  rim  horizontal.     The  rod  is  15  feet  long, 
and  a  length  of  3  feet  is  outside  of  the  bowl.     If  the  inclination  of 
the  rod  to  the  horizontal  be  30  degrees,  find  the  reactions  of  the 
bowl. 

7.  A  simple  triangular  truss,  24   feet  span  and  3  feet  deep,  is 
supported  at  the  ends  and  carries  a  load  of  3  tons  concentrated  at 
the  middle.     Find  the  stresses  in  each  member.       .  . 

175 


I76  GRAPHIC    STATICS 

8.  The  span  of  a  roof  is  16  feet ;  length  of  rafters,  9  and  12  feet, 
respectively.     The  rafters  are  spaced  2  feet  apart  and  the  roofing 
material  weighs  16  pounds  per  square  foot.     Find  the  thrust  on  each 
rafter  and  the  stress  in  the  tie  bar. 

9.  Let  a  roof  truss,  whose  general  plan  is  that  shown  in  Fig.  n, 
be  unsym metrical,  the  rafters  to  left  and  right  having  inclinations  of 
50  and  45  degrees,  respectively,  to  the  horizontal.     The  span  is  45 
feet ;  the  trusses  are  spaced  10  feet  apart ;  and  the  weight  of  the  roof 
covering  and  snow  is  40  pounds  per  square  foot  of  roof  surface.    Find 
the  stresses  in  the  members. 

10.  Find  the  stresses  in  the  members  of  this  truss  due  to  wind 
load,  assuming  the  right  end  of  the  truss  to  be  free.     Take  the  hori- 
zontal wind  pressure  as  40  pounds  per  square  foot. 

11.  A  king  post  truss  has  a  span  of  16  feet  and  a  rise  of  8  feet. 
Find  the  stresses  in  the  members  due  to  a  load  of  18,000  pounds  at 
the  middle. 

12.  A  floor  beam,  18  feet  long  and  carrying  a  uniform  load  of  180 
pounds  per  linear  foot,  is  trussed  by  rods  which  are  i|  feet  below 
the  middle  of  the  beam.     Consider  the  rods  as  jointed  in  the  middle 
of  the  beam,  and  find  the  stress  in  each  rod.     Disregard  the  weight 
of  the  beam. 

13.  A  simple  beam,  18   feet   long,  carries  a  load  of  four  tons. 
Draw  the  bending  moment  and  shearing  force  diagrams  :  (#)  when 
the  load  acts  at  the  middle  of  the  beam ;   (b)  when  it  acts  5  feet 
from  one  of  the  ends.     Disregard  the  weight  of  the  beam. 

14.  A  simple  beam  is  18  feet  long  and  loaded  with  one  ton  evenly 
distributed  along  its  span.     Find  the  bending  moment  and  the  shear- 
ing force  at  a  distance  of  6  feet  from  one  end  and  also  at  a  distance 
of  i  foot  from  the  middle.     Disregard  the  weight  of  the  beam. 

15.  On  a  common  steelyard,  a  load  of  1600  pounds  is  balanced 
by  360  pounds.     Draw  the  diagrams  of  bending  moment  and  shear- 
ing force  in  the  steelyard  bar,  when  thus  loaded. 


PROBLEMS  177 

16.  A  countershaft,  6  feet  long  between  centres  of  bearings,  car- 
ries a  driving  pulley  A,  20  inches  diameter  and  distant  i   foot  from 
the  centre  of  the  left-hand  bearing,  and  a  driven  pulley  J3,  18  inches 
diameter  and  distant  3!  feet  from  A.     In  both  cases,  the  belt-drive 
is  horizontal,  being  toward  the  front  at  A  and  to  the  rear  at  B. 
The   tension  of  the   driving  side  at  A  is  200  pounds;    at  J3,  170 
pounds.     The  weight  at  A  is  80  pounds ;  at  B,  60  pounds.     Find 
the  maximum  equivalent  bending  moment  on  the  shaft. 

17.  In  Example  14,  let  the  cranks  be  offset,  the  neutral  axis  of 
each  making  an  angle  of  120  degrees  with  the  centre  line  of  the 
shaft.     With  this  modification,  draw  the  moment  diagrams  through- 
out, as  in  Fig.  20. 

18.  In  a  4-cylinder,  triple  expansion  engine,  the  high-pressure 
and  intermediate-pressure  cylinders  are  32  and  52  inches,  respec- 
tively, in  diameter ;  the  stroke  is  48  inches ;  length  of  connecting 
rod,  96  inches ;  the  H.P.  and  I. P.  cranks  are  90  degrees  apart,  the 
H.P.  leading;  the  initial,  unbalanced  pressure  on  the  H.P.  piston 
is  200  pounds  per  square  inch  ;  on  the  I. P.  piston,  50  pounds;  the 
length  of  each  crank  pin  is  19  inches;  and  the  average  length  of 
the  three  bearings  —  one  at  each  end  of  the  section  of  the  shaft  and 
one   between   the   two   pairs  of  cranks  —  is  24  inches.     Find   the 
maximum   bending  moment  on  the  H.P.  and  I. P.  section  of  the 
shaft,  due  to   the  loads  on  the  two  pistons  and  the  corresponding 
reactions  at  the  bearings. 

19.  Assume  three  concentrated  loads,  P\,  P^  and  Ps,  spaced  at 
the  constant  distance  a  between  the  first  two  loads  and  b  between 
the  second  and  third.     Let  this  series  of  loads  cross  a  beam  of  span 
s  from  left  to  right.     Find  the  influence  line  for  the  left  reaction. 

20.  Find  the  influence  line  for  maximum  shear  for  the  series  of 
loads,  as  above. 

21.  In  Fig.  28,  let  there  be  two  coupled  locomotives  crossing  the 
plate  girder  bridge  from  left  to  right,  the  weights  and  spacing  being 
those  given  in  Example  15  and  the  locomotives  being  9  feet  apart. 
Find  the  maximum  moments  and  shears. 


178  GRAPHIC    STATICS 

22.  Let  the  5-panel  truss,  Fig.  25,  have  a  span  of  125  feet  and  a 
depth  of  26  feet.     Take  the  live  load  as  the  two  coupled  locomotives 
in  Problem  21,  followed  at  a  distance  of  5  feet  by  a  train  weighing 
5000  pounds  per  lineal  foot.     Determine  the  maximum  stresses  in 
the  members  and  find  in  which  panel  or  panels  counterbracing  is 
necessary. 

23.  Let   the  Warren  girder,   Fig.  31,  have  a  '  broken'   lower 
chord,  the  depth  at  joints  6  and  8  being  6  feet ;  at  joints  4  and  10, 
5  feet ;  and  at  joints  2  and  12,  3  feet,  4  inches.     Find  the  maximum 
live  load  stresses  in  the  members,  due  to  the  load  given  in  Example 
1 6. 

24.  Find  the  centre  of  gravity  of  the  deck  beam  whose  dimen- 
sions are  given  in  Example  20. 

25.  Find  the  centre  of  gravity  of  an  angle  with  unequal  legs, 
7x3!  inches  by  T\  inch  thick. 

26.  Find  the  rectangular  moment  of  inertia  of  the   bulb  angle 
whose  dimensions  are  given  in  Example  19.     Take  the  neutral  axis 
as  perpendicular  to  the  web  and  use  the  method  of  the  area  of  the 
equilibrium  polygon. 

27.  The  cross-section  of  a  channel  has  the  following  dimensions  : 
depth,  15  inches;  area  of  section,  9.9  square  inches;  thickness  of 
web,  0.4  inch  ;  width  of  flange,  3.4  inches  ;  thickness  of  flange,  0.9, 
tapering  to  0.4  inch ;  distance  of  centre  of  gravity  from  outside  of 
web,  0.794  inch. 

Find  the  moment  of  inertia  :  (a}  with  neutral  axis  perpendicular 
to  web  at  centre ;  (£)  with  neutral  axis  parallel  with  centre  line  of 
web. 

28.  A  cast  iron  frame  weighing  300  pounds  is  moved  across  a 
smooth  iron  floor  plate  by  a  force  P  acting  at  an  angle  of  45  degrees 
with  the  horizontal.     Take  the  coefficient  of  friction  as  0.4  and  find 
the  magnitude  of  P. 

29.  The  joint  between  the  two  sections  of  a  vertical  pump  rod  is 
made  by  a  through  key  having  one  horizontal  side  and  one  side 


PROBLEMS  179 

tapered  to  an  angle  of  31  degrees  with  the  horizontal.  In  making 
the  joint,  the  total  stress  to  be  produced  in  the  rod  is  1000  pounds. 
Using  a  coefficient  of  friction  of  0.25,  find  the  force  which  will  be 
required  to  drive  the  key  home. 

30.  The  cover  of  a  steam  engine  cylinder,  341  inches  in  diameter 
is  secured  to  the  cylinder  by  thirty-four  ij-inch  steel  studs.     Con- 
sidering friction,  what  power,  applied  to  the  nut  by  a  wrench  2  feet 
long,  will  produce  a  total  stress  in  each  nut  of  4500  pounds  ? 

31.  Find  the  efficiency,  at  the  mid-point  of  the  forward  stroke, 
of  a  double-acting,  oscillating  engine,  disregarding  the  friction  of 
the  piston  and  piston  rod.     Data:  Cylinder  diameter,  15  inches; 
stroke,  18  inches;  diameter  of  crank  pin,  2|  inches;  of  trunnion,  6 
inches  ;  unbalanced  pressure  on  piston,  100  pounds  per  square  inch  ; 
coefficient  of  friction,  0.185. 

32.  The  shaft  of  a  marine  engine   is   transmitting  4000  horse 
power  when  revolving  at  a  speed  of  125  revolutions  per  minute,  with 
the  ship  making  15  knots  per  hour.     The  thrust  shaft  hits  an  ex- 
ternal diameter  of  13!  inches.     There  are  11  thrust  collars,  each  2 
inches  wide,  working  in  a  cast  iron  bearing  lined  with  white  metal. 
Assume  —  as  was  formerly  the  practice  —  that  the  collar  bearing  is 
continuous  throughout  its  circumference,  and,  with  a  coefficient  of 
friction  of  o.i,  find  the  power  expended  in  overcoming  the  friction 
of  the  thrust  bearing. 

33.  Following  the  general  methods  given  in  Example  25,  ana- 
lyze the  action  of  the  Weston  Differential  Pulley  Block  or  chain 
hoist.     Take  the  coefficient  of  journal  friction  as  o.i,  that  of  chain 
friction  as    0.2,  and  the  ratio    between  the  diameters  of  the  two 
sheaves  in  the  upper  block  as  ^°. 

Show  the  relation  between  the  power  and  the  load  and  the 
reasons  for  the  large  loss  by  friction.  Prove  that  the  tackle  is  self- 
locking  with  regard  to  backward  motion. 

34.  In  Example  26,  let  the  pitch  diameters  of  gear  wheels  A,  B, 
and  C  be  6,   8,  and  12  inches,   respectively.     Interpose  a  lo-inch 


180  GRAPHIC    STATICS 

gear  wheel  between  wheels  B  and  C.  Taking  the  force  P  as  75 
pounds,  the  arm  L  as  8  inches,  the  arm  L'  as  15  inches,  the  coeffi- 
cient of  journal  friction  as  o.i,  and  that  of  tooth  friction  as  0.12,  find 
the  magnitude  of  the  resistance  W. 

35.  In  Fig.  52,  interpose  a  belt-tightening  pulley  to  the  left  of 
the  slack  side  (7\)  of  the  belt  and  near  the  driving  pulley  A,  thus 
altering  the  line  of  action  of  the  slack  side  and  increasing  its  arcs 
of  contact.  With  this  modification,  construct  the  force  polygons 
for  the  entire  gear  by  the  methods  used  previously. 


INDEX 


Angle  of  friction,  136 

Area,  moment  of  inertia  of,  121,  130 

partial,  centre  of  gravity  of,  1 27 
Axes  of  inertia,  parallel,  1 20 
Axis,  friction,  149 

neutral,  41 

Beam,  uniform  load,  50 
Beams,  41 
Bellcrank,  10 
Belt  gearing,  155 
Bending  moment,  43,  84 

maximum,  86 
Bending  and  twisting  moments,  49, 

61,  65 

Bending  of  ropes,  152 
Bridge  trusses,  82 
Bulb  angle,  centre  of  gravity  of,  125 

Cantilever  beams,  41 
Centre  of  gravity,  no 

of  bulb  angle,  1 25 

of  partial  area,  127 
Centrifugal  force,  load  due  to,  52 
Centroid  of  parallel  forces,  in,  113, 

124 

Chain  friction,  152 
Character  of  stress,  25 
Chord  stresses,  87 
Circle,  friction,  145 
Coefficient  of  friction,  135 
Collar,  friction  of,  142 
Cone,  friction,  137 
Continuous  beams,  41,  56 
Counterbracing,  81 


Counter  shaft,  61 
Crane  truss,  35 

pillar,  1 6 
Crank  shaft,  65 

Dead  load,  Pratt  truss,  105 

roof  truss,  26 

Warren  girder,  99 
Deck  beam,  moment  of  inertia  of, 

128 

Diagonals,  stresses  in,  87 
Diagrams,  influence,  77 

shear  and  moment,  44 

stress,  24 

Elastic  curve,  41 

Engine,  stationary,  14,  160 

Equilibrium,  conditions  of,  9 

polygon,  4 
Equivalent  moments,  49 

Force  polygon,  3 

triangle,  I 

Framed  structures,  23 
Friction,  134 

angle  of,  136 

axis  of,  149 

circle,  145 

coefficient  of,  135 

cone,  137 

force  of,  135 

of  chains,  152 

of  collars,  142 

of  gear  teeth,  158 

of  journals,  145 


181 


182 


INDEX 


Friction 

of  motion,  135 

of  pivots,  142 

of  plane  surfaces,  137 

of  rest,  135 

of  screw  jack,  164 

of  screw  threads,  139 

of  stationary  engine,  160 

radius  of,  145 

rolling,  135 

sliding,  135 

Gear  teeth,  friction  of,  158 
Girder,  stay,  56 

Warren,  98 

Gravity,  centre  of,  no,  127 
Gyration,  radius  of,  115 

Inertia,  moment  of,   115,  118,  121, 
128,  130 

Influence  diagrams,  77 
lines,  77 

Journals,  friction  of,  146 

Lines,  influence,  77 

Link  connections,  friction  of,  149 

Live  load,  73 

shear,  74,  78 

shear  maximum,  8l 

stresses,  87 

Warren  girder,  99 
Load,  dead,  27 

snow,  27 

uniform,  50,  52,  56 

wind,  30 

Loads  on  trusses,  24 
Locomotive  side  rod,  52 

wheel  loads,  89,  124 

Maximum  bending  moment, 86, 89, 98 
shear,  81,  89,  98 


Moment,  bending,  43,  84,  86,  89,  98 

diagrams,  44 

equivalent,  49 

of  inertia,  115 

of  inertia  of  area,  121,  130 

of  inertia  of  deck  beam,  128 

of     inertia      of     parallel    forces, 
118 

resisting,  43 

scale,  47 

twisting,  48 
Motion,  friction  of,  135 

Neutral  axis,  41 
Notation,  truss,  27 

Parallel  axes  of  inertia,  120 
forces,  centroid  of,  in,  113 
forces,     moment     of    inertia    of, 
118 

Pawl  and  ratchet,  1 1 

Pillar  crane,  1 6 

Pivots,  friction  of,  142 

Plane  surfaces,  friction  of,  137 

Plate  girder  bridge,  89 
girder  stresses,  87 

Polygon,  equilibrium,  4 
force,  3 

Pratt  truss,  105 

Pulley  blocks,  load  vs.  power,  167 

Radius  of  friction,  145 

of  gyration,  115 
Rachet,  11 
Ratchet-rack,  12 
Resisting  moment,  43 
Rest,  friction  of,  135 
Rolling  friction,  135 
Roof  truss,  dead  load,  26 

wind  load,  30 
Ropes,  bending  of,  152 


INDEX 


183 


Scale,  moment,  47 

Screw  jack,  friction  of,  164 

threads,  friction  of,  139 
Shaft,  centre  crank,  65 

counter,  61 
Shear  diagrams,  44 

live  load,  74,  78 

maximum,  81,  89,  98 

vertical,  42 
Sheer  legs,  20 
Simple  beam,  41 
Sliding  friction,  135 
Snow  load,  27 

Spur  gear  train,  load  vs.  power,  173 
Stationary  engine,  14,  160 
Stress,  character  of,  25 

diagrams,  24 
Supports,  free,  41 

restrained,  41 


Triangle,  force,  I 
Truss,  bridge,  82 

crane,  35 

Pratt,  105 

roof,  dead  load,  26 

roof,  wind  load,  30 

Warren,  98 

Trusses,  live  load  stresses,  87 
Twisting  and  bending,  49,  61,  65 

moments,  48 

Uniform  load,  50,  52,  56,  105 

Vertical  shear,  42 
Verticals,  stresses  in,  107 

Warren  girder,  98 
Wind  load,  30 


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Charnock,  G.  F.     Workshop  Practice.     (Westminster  Series.) 

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Church's  Laboratory  Guide.     Rewritten  by  Edward  Kinch.  .8 vo,  *2  50 

Clapperton,  G.     Practical  Papermaking 8vo,  2  50 

Clark,  C.  H.     Marine  Gas  Engines (In  Press.} 

Clark,  D.  K.     Rules,  Tables  and  Data  for  Mechanical  Engineers 

,    8vo,  5  oo 

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Tramways:  Their  Construction  and  Working 8vo,  7  50 

Clark,  J.  M.     New  System  of  Laying  Out  Railway  Turnouts . .  ' 

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Fifth  Edition 8vo,  *7  oo 

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Cleemann,  T.  M.     The  Railroad  Engineer's  Practice i2mo,  *i  50 

Clerk,  D.,  and  Idell,  F.  E.     Theory  of  the  Gas  Engine.     (Science 

Series  No.  62.) i6mo,  o  50 

Clevenger,  S.  R.     Treatise  on  the  Method  of  Government  Sur- 
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Clouth,  F.     Rubber,  Gutta-Percha,  and  Balata 8vo,  *5  oo 

Coffin,  J.  H.  C.     Navigation  and  Nautical  Astronomy i2mo,  *3  50 

Colburn,  Z.,  and  Thurston,  R.  H.     Steam  Boiler  Explosions. 

(Science  Series  No.  2.) i6mo,  o  50 

Cole,  R.  S.     Treatise  on  Photographic  Optics i2mo,  i  50 


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Collins,  J.  E.     Useful  Alloys  and  Memoranda  for  Goldsmiths, 

Jewelers i6mo,  o  50 

Constantine,  E.     Marine  Engineers,  Their    Qualifications   and 

Duties 8 vo,  *2  oo 

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Cooper,  W.  R.     Primary  Batteries 8vo,  *4  oo 

—  "  The  Electrician  "  Primers 8vo,  *5  oo 

Copperthwaite,  W.  C.     Tunnel  Shields 4to,  *g  oo 

Corey,  H.  T.     Water  Supply  Engineering 8vo  (In  Press.) 

Corfield,  W.  H.     Dwelling  Houses.     (Science  Series  No.  50.) 

i6mo,  o  50 

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Cornwall,  H.  B.  Manual  of  Blow-pipe  Analysis 8vo,  *2  50 

Courtney,  C.  F.  Masonry  Dams 8vo,  3  50 

Cowell,  W.  B.  Pure  Air,  Ozone,  and  Water i2mo,  *2  oo 

Craig,  T.  Motion  of  a  Solid  in  a  Fuel.  (Science  Series  No.  49.) 

i6mo,  o  50 

—  Wave  and  Vortex  Motion.     (Science  Series  No.  43.) .  i6mo,  o  50 

Cramp,  W.     Continuous  Current  Machine  Design 8vo,  *2  50 

Crocker,  F.  B.     Electric  Lighting.     Two  Volumes.     8vo. 

Vol.   I.     The  Generating  Plant 3  oo 

Vol.  II.     Distributing  Systems  and  Lamps 3  oo 

Crocker,  F.  B.,  and  Arendt,  M.     Electric  Motors 8vo,  *2  50 

Crocker,  F.  B.,  and  Wheeler,  S.  S.     The  Management  of  Electri- 
cal Machinery i2mo,  *i  oo 

Cross,  C.  F.,  Bevan,  E.  J.,  and  Sindall,  R.  W.     Wood  Pulp  and 

Its  Applications.  (Westminster  Series.) 8vo  (In  Press.) 

Crosskey,  L.  R.     Elementary  Prospective 8vo,  i  oo 

Crosskey,  L.  R.,  and  Thaw,  J.     Advanced  Perspective 8vo,  i  50 

Culley,  J.  L.     Theory  of  Arches.     (Science  Series  No.  87.)i6mo,  o  50 

Davenport,  C.     The  Book.     (Westminster  Series.) .8vo,  *2  oo 

Davies,  D.  C.     Metalliferous  Minerals  and  Mining 8vo,  5  oo 

Earthy  Minerals  and  Mining 8vo,  5  oo 

Davies,  E.  H.     Machinery  for  Metalliferous  Mines 8vo,  8  oo 

Davies,  F.  H.      Electric  Power  and  Traction 8vo,  *2  oo 

Dawson,  P.     Electric  Traction  on  Railways 8vo,  *g  oo 

Day,  C.    The  Indicator  and  Its  Diagrams 12  mo,  *2  oo 


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Deerr,  N.     Sugar  and  the  Sugar  Cane 8vo,  *3  oo 

Deite,  C.     Manual  of  Soapmaking.     Trans,  by  S.  T.  King.  _4to,  *5  oo 
De  la  Coux,  H.     The  Industrial  Uses  of  Water.     Trans,  by  A. 

Morris 8vo,  *4  50 

Del  Mar,  W.  A.     Electric  Power  Conductors 8vo,  *2  oo 

Denny,  G.  A.     Deep-Level  Mines  of  the  Rand 4to,  *io  oo 

—  Diamond  Drilling  for  Gold *5  oo 

De  Roos,  J.  D.  C.     Linkages.     (Science  Series  No.  47.). . .  i6mo,  o  50 

De  Varona,  A.     Sewer  Gases.     (Science  Series  No.  55.)...  i6mo,  050 

Derr,  W.  L.     Block  Signal  Operation Oblong  izmo,  *i  50 

Desaint,  A.     Three  Hundred  Shades  and  How  to  Mix  Them .  .8vo,  *io  oo 

Dibdin,  W.  J.     Public  Lighting  by  Gas  and  Electricity 8vo,  *8  oo 

—  Purification  of  Sewage  and  Water 8vo,  6  50 

Dietrich,  K.     Analysis  of  Resins,  Balsams,  and  Gum  Resins  .8vo,  *3  oo 
Dinger,  Lieut.  H.  C.     Care  and  Operation  of  Naval  Machinery 

I2D10.  *2    00 

Dixon,  D.  B.     Machinist's  and  Steam  Engineer's  Practical  Cal- 
culator   i6mo,  mor.,  i  25 

Doble,  W.  A.     Power  Plant  Construction  on  the  Pacific  Coast. 

(In  Press.) 

Dodd,  G.     Dictionary  of  Manufactures,  Mining,  Machinery,  and 

the  Industrial  Arts i2mo,  i  50 

Dorr,  B.  F.     The  Surveyor's  Guide  and  Pocket  Table-book. 

i6mo,  mor.,  2  oo' 

Down,  P.  B.     Handy  Copper  Wire  Table i6mo,  *i  oo 

Draper,   C.    H.     Elementary   Text-book   of   Light,    Heat   and 

Sound 1 2mo,  i  oo 

Heat  and  the  Principles  of  Thermo-dynamics i2mo,  i  50 

Duckwall,  E.  W.    Canning  and  Preserving  of  Food  Products. 8 vo,  *5  oo 

Dumesny,  P.,  and  Noyer,  J.     Wood  Products,  Distillates,  and 

Extracts 8vo,  *4  50 

Duncan,  W.  G.,  and  Penman,  D.     The  Electrical  Equipment  of 

Collieries 8vo,  *4  oo 

Duthie,    A.    L.     Decorative    Glass    Processes.     (Westminster 

Series) ..8vo,  *2  oo 

Dyson,  S.  S.     Practical  Testing  of  Raw  Materials 8vo,  *5  oo 

Eccles,  R.  G.,  and  Duckwall,  E.  W.     Food  Preservatives 8vo,  i  oo 

paper,  o  50 


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Eddy,  H.  T.     Researches  in  Graphical  Statics 8vo,  i  50 

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Edgcumbe,  K.     Industrial  Electrical  Measuring  Instruments . 

8vo,  *2  50 

Eissler,  M.     The  Metallurgy  of  Gold 8vo,  7  50 

—  The  Hydrometallurgy  of  Copper 8vo,  *4  50 

—  The  Metallurgy  of  Silver 8vo,  4  oo 

The  Metallurgy  of  Argentiferous  Lead 8vo,  5  oo 

Cyanide  Process  for  the  Extraction  of  Gold 8vo,  3  oo 

A  Handbook  of  Modern  Explosives 8vo,  5  oo 

Ekin,  T.  C.     Water  Pipe  and  Sewage  Discharge  Diagrams . .  folio,  *3  oo 
Eliot,  C.  W.,  and  Storer,  F.  H.    Compendious  Manual  of  Qualita- 
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Elliot,  Major  G.  H.     European  Light-house  Systems 8vo,  5  oo 

Ennis,  Wm.  D.     Linseed  Oil  and  Other  Seed  Oils   8vo,  *4  oo 

Applied  Thermodynamics 8 vo,  *4  50 

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Erfurt,  J.     Dyeing  of  Paper  Pulp.     Trans,  by  J.  Hubner . . .  8vo,  *7  50 

Erskine -Murray,  J.     A  Handbook  of  Wireless  Telegraphy.. 8 vo,  *3  50 

Evans,  C.  A.     Macadamized  Roads (In  Press.) 

Ewing,  A.  J.     Magnetic  Induction  in  Iron 8vo,  *4  oo 

Fairie,  J.     Notes  on  Lead  Ores i2mo,  *i  oo 

—  Notes  on  Pottery  Clays i2mo,  *i  50 

Fairley,  W.,  and  Andre,  Geo.  J.     Ventilation  of  Coal  Mines. 

(Science  Series  No.  58.) i6mo,  o  50 

Fairweather,  W.  C.     Foreign  and  Colonial  Patent  Laws  . .  .8vo,  *3  oo 
Fanning,    T.    T.     Hydraulic   and   Water-supply    Engineering. 

8vo,  *s  oo 
Fauth,  P.     The  Moon  in  Modern  Astronomy.     Trans,  by  J. 

McCabe 8vo,  *2  oo 

Fay,  I.  W.     The  Coal-tar  Colors 8vo  (In  Press.) 

Fernbach,  R.  L.    Glue  and  Gelatine 8vo,  *3  oo 

Fischer,  E.     The  Preparation  of  Organic  Compounds.     Trans. 

by  R.  V.  Stanford i2mo,  *i  25 

Fish,  J.  C.  L.     Lettering  of  Working  Drawings Oblong  8vo,  i  oo 

Fisher,  H.  K.  C.,  and  Darby,  W.  C.     Submarine  Cable  Testing. 

8vo,  *3  50 

Fiske,  Lieut.  B.  A.     Electricity  in  Theory  and  Practice *8vo,  2  50 


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Fleischmann,  W.     The  Book  of  the  Dairy.     Trans,  by  C.  M. 

Aikman 8 vo,  4  oo 

Fleming,    J.    A.     The    Alternate-current    Transformer.     Two 

Volumes 8  vo, 

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Vol.  II.     The  Utilization  of  Induced  Currents *5  oo 

Centenary  of  the  Electrical  Current 8vo,  *o  50 

Electric  Lamps  and  Electric  Lighting 8 vo,  *3  oo 

Electric  Laboratory  Notes  and  Forms 4to,  *5  oo 

A  Handbook  for  the  Electrical  Laboratory  and  Testing 

Room.     Two  Volumes 8vo,  each,  *5  oo 

Fluery,  H.     The  Calculus  Without   Limits  or    Infinitesimals. 

Trans,  by  C.  O.  Mailloux (In  Press.) 

Flynn,  P.  J.     Flow  of  Water.     (Science  Series  No.  84.). . .  i6mo,  o  50 

Hydraulic  Tables.     (Science  Series  No.  66.) i6mo,  o  50 

Foley,  N.     British  and  American  Customary  and  Metric  Meas- 
ures   folio,  *3  oo 

Foster,    H.    A.     Electrical    Engineers'     Pocket-book.     (Sixth 

Edition.} i2mo,  leather,  5  oo 

Foster,   Gen.   J.   G.     Submarine   Blasting   in   Boston    (Mass.) 

Harbor 4to,  3  50 

Fowle,  F.  F.     Overhead  Transmission  Line  Crossings  .. .  .i2mo,  *i  50 
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Fox,  W.  G.     Transition  Curves.     (Science  Series  No.  no. ).i6mo,  050 
Fox,  W.,  and  Thomas,  C.  W.     Practical  Course  in  Mechanical 

Drawing i2mo,  i  25 

Foye,  J.  C.     Chemical  Problems.     (Science  Series  No.  6g.).i6mo,  o  50 

—  Handbook  of  Mineralogy.     (Science  Series  No.  86.). .  i6mo,  o  50 

Francis,  J.  B.     Lowell  Hydraulic  Experiments 4to,  15  oo 

Frye,  A.  I.     Civil  Engineers'  Pocket-book (In  Press.) 

Fuller,  G.  W.     Investigations  into  the  Purification  of  the  Ohio 

River 4to,  *io  oo 

Furnell,  J.     Paints,  Colors,  Oils,  and  Varnishes 8vo,  *i  oo 

Gant,  L.  W.     Elements  of  Electric  Traction 8vo,  *2  50 

Garcke,  E.,  and  Fells,  J.  M.     Factory  Accounts 8vo,  3  oo 

Garforth,  W.  E.     Rules  for  Recovering  Coal  Mines  after  Explo- 
sions and  Fires i2mo,  leather,  i  50 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     13 

Gaudard,  J.     Foundations.     (Science  Series  No.  34.) i6mo,  o  50 

Gear,  H.  B.,  and  Williams,  P.  F.     Electric  Central  Station  Dis- 
tributing Systems. . . 8vo  (In  Preparation.} 

Geerligs,  H.  C.  P.     Cane  Sugar  and  Its  Manufacture 8vo,  *5  oo 

Geikie,  J.     Structural  and  Field  Geology 8vo,  *4  oo 

Gerber,  N.     Analysis  of  Milk,  Condensed  Milk,  and  Infants' 

Milk-Food .8vo,  i  25 

Gerhard,  W.  P.     Sanitation,  Water-supply  and  Sewage  Disposal 

of  Country  Houses i2mo,  *2  oo 

—  Gas  Lighting.     (Science  Series  No.  in.) i6mo,  o  50 

—  Household  Wastes.     (Science  Series  No.  97,) i6mo,  o  50 

—  House  Drainage.     (Science  Series  No.  63.) i6mo,  o  50 

—  Sanitary  Drainage  of  Buildings.     (Science  Series  No.  93.) 

i6mo,  o  50 

Gerhardi,  C.  W.  H.     Electricity  Meters 8vo,  *4  oo 

Geschwind,  L.     Manufacture  of  Alum  and  Sulphates,     Trans. 

by  C.  Salter 8vo,  *$  oo 

Gibbs,  W.  E.     Lighting  by  Acetylene i2mo,  *i  50 

—  Physics  of  Solids  and  Fluids.     (Carnegie  Technical  Schools 

Text-books.) *i  50 

Gibson,  A.  H.     Hydraulics  and  Its  Application 8vo,  *5  oo 

—  Water  Hammer  in  Hydraulic  Pipe  Lines . . ., i2mo,  *2  oo 

Gilbreth,  F.  B.     Motion  Study.     A  Method  for  Increasing  the 

Efficiency  of  the  Workman i2mo,  *2  oo 

Gillmore,  Gen.  Q.  A.     Limes,  Hydraulic  Cements  and  Mortars. 

8vo,  4  oo 

•  Roads,  Streets,  and  Pavements i2mo,  2  oo 

Golding,  H.  A*    The  Theta-Phi  Diagram i2mo,  *i  25 

Goldschmidt,  R.     Alternating  Current  Commutator  Motor  .8vo,  *3  oo 

Goodchild,  W.     Precious  Stones.     (Westminster  Series.). .  .8vo,  *2  oo 

Goodeve,  T.  M.     Textbook  on  the  Steam-engine i2mo,  2  oo 

Gore,  G.     Electrolytic  Separation  of  Metals 8vo,  *3  50 

Gould,  E.  S.     Arithmetic  of  the  Steam-engine ,. i2mo,  i  oo 

—  Calculus.     (Science  Series  No.  112.) i6mo,  o  50 

—  High  Masonry  Dams.     (Science  Series  No.  22.) i6mo,  o  50 

Practical  Hydrostatics  and  Hydrostatic  Formulas.     (Science 

Series.) i6mo,  o  50 

Grant,  J.  Brewing  and  Distilling.     (Westminster  Series.)  8vo  (In  Press.) 

Gray,  J,    Electrical  Influence  Machines i2mo,  2  oo 


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Greenwood,  E.     Classified  Guide  to  Technical  and  Commercial 

Books 8vo,  *3  oo 

Gregorius,  R.     Mineral  Waxes.     Trans,  by  C.  Salter . ....  i2mo,  *3  oo 

Griffiths,  A.  B.     A  Treatise  on  Manures i2mo,  3  oo 

—  Dental  Metallurgy 8vo,  *^  50 

Gross,  E.     Hops 8 vo,  *4  50 

Grossman,  J.     Ammonia  and  its  Compounds i2mo,  *i  25 

Groth,  L.  A.     Welding  and  Cutting  Metals  by  Gases  or  Electric- 
ity  8vo,  *3  oo 

Grover,  F.     Modern  Gas  and  Oil  Engines 8vo,  *2  oo 

Gruner,  A.     Power-loom  Weaving 8vo,  *3  oo 

Guldner,    Hugo.      Internal-Combustion    Engines.      Trans,    by 

H.  Diedrichs 4to,  *io  oo 

Gunther,  C.  O.     Integration i2mo,  *i  25 

Gurden,  R.  L.     Traverse  Tables folio,  half  mor.  7  50 

Guy,  A.  E.     Experiments  on  the  Flexure  of  Beams 8vo,  *i  25 

Haeder,  H.     Handbook  on  the  Steam-engine.     Trans,  by  H.  H. 

P.  Powles i2mo,  3  oo 

Hainbach,  R.     Pottery  Decoration.     Trans,  by  C.  Slater.    i2mo,  *3  oo 

Hale,  W.  J.     Calculations  of  General  Chemistry i2mo,  *i  oo 

Hall,  C.  H.     Chemistry  of  Paints  and  Paint  Vehicles i2mo,  *2  oo 

Hall,  R.  H.     Governors  and  Governing  Mechanism i2mo,  *2  oo 

Hall,  W.  S.     Elements  of  the  Differential  and  Integral  Calculus 

8vo,  *2  25 

—  Descriptive  Geometry 8vo  volume  and  4to  atlas,  *3  50 

Haller,  G.  F.,  and  Cunningham,  E.  T.    The  Tesla  Coil i2mo,  *i  25 

Halsey,  F.  A.     Slide  Valve  Gears i2mo,  i  50 

—  The  Use  of  the  Slide  Rule.     (Science  Series.) i6mo,  o  50 

—  Worm  and  Spiral  Gearing.     (Science  Series.). ......  i6mo,  o  50 

Hamilton,  W.  G.     Useful  Information  for  Railway  Men..i6mo,  i  oo 

Hammer, W.  J.     Radium  and  Other  Radioactive  Substances,  8 vo,  *i  oo 

Hancock,  H.     Te  xtbook  of  Mechanics  and  Hydrostatics. ...  .8vo,  i  50 

Hardy,  E.     Elementary  Principles  of  Graphic  Statics i2mo,  *i  50 

Harper,  W.  B.     Utilization  of  Wood  Waste  by  Distillation. .  4to,  *3  oo 

Harrison,  W.  B.     The  Mechanics'  Tool-book i2mo,  i  50 

Hart,  J.  W.     External  Plumbing  Work 8vo,  *3  oo 

—  Hints  to  Plumbers  on  Joint  Wiping 8vo,  *3  oo 

—  Principles  of  Hot  Water  Supply 8vo,  *3  oo 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     15 

Hart,  J.  W.     Sanitary  Plumbing  and  Drainage 8vo,  *3  oo 

Haskins,  C.  H.     The  Galvanometer  and  Its  Uses i6mo,  i  50 

Hatt,  J.  A.  H.     The  Colorist square  i2mo,  *i  50 

Hausbrand,  E.     Drying  by  Means  of  Air  and  Steam.     Trans. 

by  A.  C.  Wright i2mo,  *2  oo 

—  Evaporing,  Condensing  and  Cooling  Apparatus.     Trans. 

by  A.  C.  Wright 8vo,  *5  oo 

Hausner,  A.  Manufacture  of  Preserved  Foods  and  Sweetmeats. 

Trans,  by  A.  Morris  and  H.  Robson. 8vo,  *3  oo 

Hawke,  W.  H.  Premier  Cipher  Telegraphic  Code 4to,  *5  oo 

—  100,000  Words  Supplement  to  the  Premier  Code 4to,  *5  oo 

Hawkesworth,  J.     Graphical  Handbook  for  Reniforced  Concrete 

Design. 4to,  *2  50 

Hay,  A.     Alternating  Currents 8vo,  *2  50 

—  Principles  of  Alternate-current  Working i2mo,  2  oo 

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—  Continuous  Current  Engineering 8vo,  *2  50 

Heap,  Major  D.  P.     Electrical  Appliances 8vo,  2  oo 

Heaviside,    O.     Electromagnetic    Theory.     Two    volumes. 

8vo,  each,  *5  oo 

Heck,  R.  C.  H.     Steam-Engine  and  Other  Steam  Motors.     Two 
Volumes. 

Vol.   I.     Thermodynamics  and  the  Mechanics 8vo,  *3  50 

Vol.  II.     Form,  Construction  and  Working 8vo,  *5  oo 

Abridged  edition  of  above  volumes  (Elementary) 

8vo  (In  Preparation) 

—  Notes  on  Elementary  Kinematics 8vo,  boards,  *i  oo 

—  Graphics  of  Machine  Forces , 8vo,  boards,  *i  oo 

Hedges,  K.     Modern  Lightning  Conductors 8vo,  3  oo 

Heermann,  P.     Dyers'  Materials.      Trans,  by  A.  C.  Wright. 

i2mo,  *2  50 
Hellot,  Macquer  and  D'Apligny.  Art  of  Dyeing  Wool,  Silk  and 

Cotton 8vo,  *2  oo 

Henrici,  O.  Skeleton  Structures. 8vo,  i~  50 

Hermann,  F.  Painting  on  Glass  and  Porcelain 8vo,  *3  50 

Hermann,  G.  The  Graphical  Statics  of  Mechanism.  Trans. 

by  A.  P.  Smith i2mo,  2  oo 

Herzfeld,  J.  Testing  of  Yarns  and  Textile  Fabrics 8vo,  *3  50 

Hildebrandt,  A.  Airships,  Past  and  Present 8vo,  *3  50 


16    D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 


Hildenbrand,  B.  W.     Cable-Making.     (Science  Series  No..  32.) 

i6mo,  b  50 

Hildrich,  H.     Short  History  of  Chemistry (In  Press.) 

Hill,  J.  W.     The  Purification  of  Public  Water  Supplies.     New 

Edition (In  Press.) 

Interpretation  of  Water  Analysis (In  Press.) 

Hiroi,  I.     Plate  Girder  Construction.     (Science  Series  No.  95.) 

i6mo,  o  50 

Statically-Indeterminate  Stresses I2mo,  *2  oo 

Hirshfeld,    C.    F.      Engineering     Thermodynamics.     (Science 

Series.). i6mo,  o  50 

Hobart,  H.  M.     Heavy  Electrical  Engineering 8vo,  *4  50 

Electricity 8vo,  *2  oo 

Electric  Trains 8vo,  *2  50 

Hobbs,  W.  R.  P.     The  Arithmetic  of  Electrical  Measurements 

1 2 mo,  o  50 

Hoff,  J.  N.     Paint  and  Varnish  Facts  and  Formulas i2mo,  *i  50 

Hoflf,  Com.W.  B.  The  Avoidance  of  Collisions  at  Sea.  i6mo,  mor.,     o  75 

Hole,  W.     The  Distribution  of  Gas 8vo,  *7  50 

Holley,  A.  L.     Railway  Practice folio,  12  oo 

Holmes,  A.  B.     The  Electric  Light  Popularly  Explained. 

i2mo,  paper,  o  50 

Hopkins,  N.  M.     Experimental  Electrochemistry 8vo,  *3  oo 

Model  Engines  and  Small  Boats I2mo,  i  25 

Hopkinson,  J.,  Shoolbred,  J.  N.,  and  Day,  R.  E.     Dynamic 

Electricity.     (Science  Series  No.  71.) i6mo,  o  50 

Horner,  J.     Engineers'  Turning 8vo,  *3  50 

Metal  Turning I2mo,  i  50 

Toothed  Gearing i2mo,  2  25 

Houghton,  C.  E.     The  Elements  of  Mechanics  of  Materials.  i2mo,  *2  oo 

Houllevigue,  L.     The  Evolution  of  the  Sciences 8vo,  *2  oo 

Howe,  G.     Mathematics  for  the  Practical  Man i2mo,  *i  25 

Howorth,  J.     Repairing  and  Riveting  Glass,  China  and  Earthen- 
ware  8vo,  paper,  *o  50 

Hubbard,  E.     The  Utilization  of  Wood- waste 8vo,  *2  50 

Humber,  W.     Calculation  of  Strains  in  Girders i2mo,  2  50 

Humphreys,    A.    C.     The    Business   Features    of   Engineering 

Practice 8vo,  *i  25 

Hurst,  G.  H.    Handbook  of  the  Theory  of  Color, 8vo,  *2  50 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     17 

Hurst,  G.H.     Dictionary  of  Chemicals  and  Raw  Products.. 8vo,  *3  oo 

—  Lubricating  Oils,  Fats  and  Greases. 8vo,  *3  oo 

—  Soaps 8vo,  *5  oo 

—  Textile  Soaps  and  Oils .  8vo,  *2  50 

Hurst,  H.  E.,  and  Lattey,  R.  T.     Text-book  of  Physics 8vo,  *3  oo 

Hutchinson,  R.  W.,  Jr.     Long  Distance  Electric  Power  Trans- 
mission  i2mo,  *3  oo 

Hutchinson,  R.  W.,  Jr.,  and  Ihlseng,  M.  C.  Electricity  in 

Mining i2mo  (In  Press.) 

Hutchinson,  W.  B.  Patents  and  How  to  Make  Money  Out  of 

Them i2mo,  i  25 

Hutton,  W.  S.     Steam-boiler  Construction 8vo,  6  oo 

—  Practical  Engineer's  Handbook 8vo,  7  oo 

—  The  Works'  Manager's  Handbook 8vo,  6  oo 

Hyde,  E.  W.    Skew  Arches.     (Science  Series  No.  15.)..  •  •  i6mo,  o  50 

Induction  Coils.     (Science  Series  No.  53.). i6mo,  o  50 

Ingle,  H.     Manual  of  Agricultural  Chemistry 8vo,  *3  oo 

Innes,  C.  H.     Problems  in  Machine  Design i2mo,  *2  oo 

—  Air  Compressors  and  Blowing  Engines i2mo,  *2  oo 

—  Centrifugal  Pumps i2mo,  *2  oo 

—  The  Fan i2mo,  *2  oo 

Isherwood,  B.  F.     Engineering  Precedents  for  Steam  Machinery 

8vo,  2  50 

Ivatts,  E.  B.     Railway  Management  at  Stations 8vo,  *2  50 

Jacob,  A.,  and  Gould,  E.  S.     On  the  Designing  and  Construction 

of  Storage  Reservoirs.     (Science  Series  No.  6.).  .  i6mo,  o  50 

Jamieson,  A.     Text  Book  on  Steam  and  Steam  Engines. .  .   8vo,  3  oo 

—  Elementary  Manual  on  Steam  and  the  Steam  Engine  .  1 2mo,  i  50 
Jannettaz,  E.     Guide  to  the  Determination  of  Rocks.     Trans. 

by  G.  W.  Plympton ." i2mo,  i  50 

Jehl,  F.     Manufacture  of  Carbons 8vo,  *4  oo 

Jennings,    A.    S.     Commercial   Paints   and   Painting.     (West- 
minster Series.) 8vo  (In  Press.) 

Jennison,  F.  H.     The  Manufacture  of  Lake  Pigments 8vo,  *3  oo 

Jepson,  G.     Cams  and  the  Principles  of  their  Construction...  8vo,  *i  50 

—  Mechanical  Drawing 8vo  '(In  Preparation.) 


18     D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Johnson,  G.  L.     Photographic  Optics  and  Color  Photography 

8vo,  *3  oo 
Johnson,  W.  H.  The  Cultivation  and  Preparation  of  Para 

Rubber '. 8vo,  *3  oo 

Johnson,  W.  McA.     The  Metallurgy  of  Nickel (In  Preparation.) 

Johnston,  J.  F.  W.,  and  Cameron,  C.     Elements  of  Agricultural 

Chemistry  and  Geology i2mo,  2  60 

Joly,  J.     Radioactivity  and  Geology i2mo,  *3  oo 

Jones,  H.  C.     Electrical  Nature  of  Matter  and  Radioactivity 

I2H1O,  2    OO 

Jones,  M.  W.     Testing  Raw  Materials  Used  in  Paint. ....  i2mo,  *2  oo 

Jones,  L.,  and  Scard,  F.  I.     Manufacture  of  Cane  Sugar 8vo,  *s  oo 

Joynson,  F.  H.     Designing  and  Construction  of  Machine  Gear- 
ing.  8vo,  2  oo 

Jiiptner,  H.  F.  V.     Siderology :  The  Science  of  Iron 8vo,  *5  oo 

Kansas  City  Bridge 4to,  6  oo 

Kapp,  G.  Alternate  Current  Machinery.  (Science  Series  No. 

96.) i6mo,  o  50 

Dynamos,  Motors,  Alternators  and  Rotary  Converters. 

Trans,  by  H.  H.  Simmons 8vo,  4  oo 

—  Electric  Transmission  of  Energy . .. i2mo,  3  50 

Keim,  A.  W.     Prevention  of  Dampness  in  Buildings 8vo,  *2  oo 

Keller,  S.  S.     Mathematics  for  Engineering  Students. 

i2mo,  half  leather, 

—  Algebra  and  Trigonometry,  with  a  Chapter  oil  Vectors. ...  *i  75 

—  Special  Algebra  Edition *i  oo 

—  Plane  and  Solid  Geometry *i  25 

—  Analytical  Geometry  and  Calculus *2  oo 

Kelsey,  W.    R.      Continuous-current    Dynamos  and  Motors. 

8vo,  *2  50 
Kemble,  W.  T.,  and  Underbill,  C.  R.  The  Periodic  Law  and  the 

Hydrogen  Spectrum 8vo,  paper,  *o  50 

Kemp,  J.  F.  Handbook  of  Rocks 8vo,  *i  50 

Kendall,  E.  Twelve  Figure  Cipher  Code. 4to,  *i$  oo 

Kennedy,  A.  B.  W.,  and  Thurston,  R.  H.  Kinematics  of 

Machinery.  (Science  Series  No.  54.) i6mo,  o  50 

Kennedy,  A.  B.  W.,  Unwin,  W.  C.,  and  Idell,  F.  E.  Compressed 

Air.     (Science  Series  No.  106.) i6mo,  o  50 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     19 

Kennedy,  R.     Modern  Engines  and  Power   Generators.     Six 

Volumes 4to,  15  oo 

Single  Volumes each,  3  oo 

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Single  Volumes each,  3  50 

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Kennelly,  A.  E.    Electro-dynamic  Machinery 8vo,  i  50 

Kent,  W.    Strength  of  Materials.     (Science  Series  No.  41.).  i6mo,  o  50 

Kershaw,  J.  B.  C.     Fuel,  Water  and  Gas  Analysis 8vo,  *2  50 

—  Electrometallurgy.     (Westminster  Series.) 8vo,  *2  oo 

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Kingdon,  J.  A.     Applied  Magnetism 8vo,  *3  oo 

Kinzbrunner,  C.     Alternate  Current  Windings. 8vo,  *i  50 

—  Continuous  Current  Armatures 8vo,  *i  50 

—  Testing  of  Alternating  Current  Machines 8vo,  *2  oo 

Kirkaldy,    W.    G.     David    Kirkaldy's    System    of    Mechanical 

Testing 4to,  10  oo 

Kirkbride,  J.     Engraving  for  Illustration 8vo,  *i  50 

Kirkwood,  J.  P.     Filtration  of  River  Waters 4to,  7  50 

Klein,  J.  F.     Design  of  a  High  speed  Steam-engine 8vo,  *5  oo 

—  Physical  Significance  of  Entropy 8vo,  *i  50 

Kleinhans,  F.  B.     Boiler  Construction 8vo,  3  oo 

Knight,  Capt.  A.  M.     Modern  Steamship 8vo,  *7  50 

Half  Mor.  *Q  oo 

Knox,  W.  F.     Logarithm  Tables (In  Preparation.) 

Knott,  C.  G.,  and  Mackay,  J.  S.     Practical  Mathematics.  .  .8vo,  2  oo 

Koester,  F.     Steam-Electric  Power  Plants 4to,  *5  oo 

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Koller,  T.     The  Utilization  of  Waste  Products 8vo,  *3  50 

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Krauch,  C.     Testing  of  Chemical  Reagents.     Trans,  by  J.  A. 

Williamson  and  L.  W.  Dupre 8vo,  *3  oo 

Lambert,  T.     Lead  and  its  Compounds 8vo,  *3  50 

—  Bone  Products  and  Manures 8vo,  *3  oo 

Lamborn,  L.  L.     Cottonseed  Products 8vo,  *3  oo 

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Lamprecht,  R.     Recovery  Work  After  Pit  Fires.      Trans,  by 

C.  Salter 8vo,  *4  oo 


20     D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATA] 

Lanchester,  F.  W.     Aerial  Flight.     Two  Volumes.     8vo. 

Vol.    I.     Aerodynamics. /  .  *6  oo 

Vol.  II.     Aerodonetics , .  .  .  *6  oo 

Lamer,  E.  T.     Principles  of  Alternating  Currents jimo,  *i  25 

Larrabee,   C.   S.     Cipher   and   Secret   Letter   and   Telegraphic 

Code i6mo,  o  60 

La  Rue,  B.  F.     Swing  Bridges.     (Science  Series  No.  107.) .  i6mo,  o  50 
Lassar-Cohn,  Dr.     Modern  Scientific  Chemistry.     Trans,  by  M. 

M.  Pattison  Muir i2mo,  *2  oo 

Latimer,  L.  H.,  Field,  C.  J.,  and  Howell,  J.  W.     Incandescent 

Electric  Lighting.     (Science  Series  No.  57.) i6mo,  o  50 

Latta,  M.  N.     Handbook  of  American  Gas-Engineering  Practice. 

8vo,  *4  50 

—  American  Producer  Gas  Practice 4to,  *6  oo 

Leask,  A.  R.     Breakdowns  at  Sea i2mo,  2  oo 

—  Triple  and  Quadruple  Expansion  Engines i2mo,  2  oo 

—  Refrigerating  Machinery , i2mo,  2  oo 

Lecky,  S.  T.  S.     "  Wrinkles  "  in  Practical  Navigation 8vo,  *8  oo 

Le  Doux,  M.     Ice-Making  Machines.     (Science  Series  No.  46.) 

i6mo,  o  50 
Leeds,  C.  C.    Mechanical  Drawing  for  Trade  Schools .  oblong,  4to, 

High  School  Edition *i  25 

Machinery  Trades  Edition *2  oo 

Lefgvre,  L.     Architectural  Pottery.     Trans,  by  H.  K.  Bird  and 

W.  M-.  Binns 4to,  *7  50 

Lehner,  S.     Ink  Manufacture.     Trans,  by  A.  Morris  and  H. 

Robson 8vo,  *2  50 

Lemstrom,  S.     Electricity  in  Agriculture  and  Horticulture. 

8vo,  *i  50 
Le  Van,  W.  B.     Steam-Engine  Indicator.     (Science  Series  No. 

78.) .  .  i6mo,  o  50 

Lewes,  V.  B.     Liquid  and  Gaseous  Fuels.     (Westminster  Series.) 

8vo,  *2  oo 

Lieber,  B.  F.     Lieber's  Standard  Telegraphic  Code 8vo,  *io  oo 

—  Code.     German  Edition 8vo,  *io  oo 

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French  Edition 8vo,  *io  oo 

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Lieber's  Appendix folio,  *i$  oo 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG   21 

Lieber,    B.  F.     Handy  Tables. 4to,     *2  50 

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Shippers'  Blank  Tables 8vo,  *i$  oo 

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Livermore,  V.  P.,  and  Williams,  J.     How  to  Become  a  Com- 
petent Motorman i2mo,  *i  oo 

Livingstone,  R.     Design  and  Construction  of  Commutators.  8vo,  *2  25 

Lobben,  P.     Machinists'  and  Draftsmen's  Handbook  ......  8vo,  2  50 

Locke,  A.  G.  and  C.  G.     Manufacture  of  Sulphuric  Acid 8vo,  10  oo 

Lockwood,  T.  D.     Electricity,  Magnetism,  and  Electro-teleg- 
raphy  8vo,  2  50 

—  Electrical  Measurement  and  the  Galvanometer i2mo,  i  50 

Lodge,  0.  J.     Elementary  Mechanics i2mo,  i  50 

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Lord,  R.  T.     Decorative  and  Fancy  Fabrics 8vo,  *3  50 

Loring,  A.  E.     A  Handbook  of  the  Electromagnetic  Telegraph. 

(Science  Series  No.  39) i6mo,       o  50 

Loewenstein,  L.  C.,  and  Crissey,  C.  P.     Centrifugal  Pumps.     (In  Press.) 
Lucke,  C.  E.     Gas  Engine  Design 8vo,     *3  oo 

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2  vols (In  Preparation.) 

—  Power  Plant  Papers.     Form  I.     The  Steam  Power  Plant 

paper,     *i  50 
Lunge,  G.     Coal-tar  Ammonia.     Two  Volumes 8vo,  *i$  oo 

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8vo, 

Vol.    I.     Sulphuric  Acid.     In  two  parts *i$  oo 

Vol.  II.     Salt  Cake,  Hydrochloric  Acid  and  Leblanc  Soda. 

In  two  parts *  15  oo 

Vol.  III.     Ammonia  Soda *  15  oo 

—  Technical  Chemists'  Handbook i2mo,  leather,     *3  50 

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Keane.     In  collaboration  with  the  corps  of  specialists. 

Vol.    I.     In  two  parts 8vo,  *  15  oo 

Vols.  II  and  III (In  Preparation.) 

Lupton,  A.,  Parr,  G.  D.  A.,  and  Perkin,  H.     Electricity  as  Applied 

to  Mining ,. .... ... ... ... . .  .8vo,     *4  50 

Luquer,  L.  M.    Minerals  in  Rock  Sections 8vo,     *i  50 


22    D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLK  CATALOG 

Macewen,  H.  A.     Food  Inspection 8vo,  *2  50 

Mackenzie,  N.  F.     Notes  on  Irrigation  Works 8vo,  *2  50 

Mackie,  J.     How  to  Make  a  Woolen  Mill  Pay 8vo,  *2  oo 

Mackrow,    C.     Naval    Architect's    and    Shipbuilder's    Pocket- 
book i6mo,  leather,  5  oo 

Maguire,  Capt.  E.     The  Attack  and  Defense  of  Coast  Fortifica- 
tions  8vo,  2  50 

Maguire,  Wm.  R.     Domestic  Sanitary  Drainage  and  Plumbing 

8vo,  4  oo 
Mallet,    A.     Compound    Engines.     Trans,    by    R.    R.    Buel. 

(Science  Series  No.  10.) i6mo, 

Mansfield,  A.  N.     Electro-magnets.     (Science  Series  No.  64) 

i6mo,  o  50 
Marks,  E.  C.  R.     Construction  of  Cranes  and  Lifting  Machinery 

12010,  *i  50 

Construction  and  Working  of  Pumps i2mo,  *i  50 

Manufacture  of  Iron  and  Steel  Tubes i2mo,  *2  oo 

Mechanical  Engineering  Materials i2mo,  *i  oo 

Marks,  G.  C.     Hydraulic  Power  Engineering 8vo,  3  50 

—  Inventions,  Patents  and  Designs i2mo,  *i  oo 

Markham,  E.  R.     The  American  Steel  Worker i2mo,  2  50 

Marlow,  T.  G.     Drying  Machinery  and  Practice 8vo,  *5  oo 

Marsh,  C.  F.     Concise  Treatise  on  Reinforced  Concrete.. .  .8vo,  *2  50 

Marsh,  C.  F.,  and  Dunn,  W.     Reinforced  Concrete 4to,  *5  oo 

—  Manual  of  Reinforced  Concrete  and  Concrete  Block  Con- 

struction  i6mo,  mor.,  *2  50 

Marshall,  W.J.,  and  Sankey,  H.  R.  Gas  Engines.  (Westminster 

Series.) 8vo,  *2  oo 

Massie,  W.  W.,  and  Underbill,  C.  R.  Wireless  Telegraphy  and 

Telephony i2mo,  *i  oo 

Matheson,  D.  Australian  Saw-Miller's  Log  and  Timber  Ready 

Reckoner i2mo,  leather,  i  50 

Mathot,  R.  E.  Internal  Combustion  Engines 8vo,  *6  oo 

Maurice,  W.  Electric  Blasting  Apparatus  and  Explosives  ..8 vo,  *3  50 

—  Shot  Firer's  Guide 8vo,  *  i  50 

Maxwell,  J.  C.     Matter  and  Motion.     (Science  Series  No.  36.) 

i6mo,  o  50 
Maxwell,  W.  H.,  and  Brown,  J,  T.     Encyclopedia  of  Municipal 

and  Sanitary  Engineering 4to,  *io  oo 

Mayer,  A.  M.     Lecture  Notes  on  Physics 8vo,  2  oo 


D.  VAiN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     23 

McCullough,  R.  S.     Mechanical  Theory  of  Heat 8vo,       3  50 

Mclntosh,  J.  G.     Technology  of  Sugar 8vo,     *4  50 

—  Industrial  Alcohol 8vo,     *3  oo 

Manufacture  of  Varnishes  and  Kindred  Industries.    Three 

Volumes.     8vo. 

Vol.  I.     Oil  Crushing,  Refining  and  Boiling *3  50 

Vol.  II.     Varnish  Materials  and  Oil  Varnish  Making *4  oo 

Vol.  in (In  Preparation.) 

McKnight,  J.   D.,  and  Brown,   A.   W.     Marine   Multitubular 

Boilers *i  50 

McMaster,  J.  B.     Bridge  and  Tunnel  Centres.     (Science  Series 

No.  20.) i6mo,       o  50 

McMechen,  F.  L.     Tests  for  Ores,  Minerals  and  Metals.. .  i2mo,     *i  oo 

McNeill,  B.     McNeill's  Code 8vo,     *6  oo 

McPherson,  J.  A.     Water-works  Distribution 8vo,       2  50 

Melick,  C.  W.     Dairy  Laboratory  Guide 121110,     *i  25 

Merck,  E.  Chemical  Reagents ;  Their  Purity  and  Tests ....  8vo,  *i  50 
Merritt,  Wm.  H.  Field  Testing  for  Gold  and  Silver .  i6mo,  leather,  i  50 
Meyer,  J.  G.  A.,  and  Pecker,  C.  G.  Mechanical  Drawing  and 

Machine  Design 4to,       5  oo 

Michell,  S.     Mine  Drainage 8vo,     10  oo 

Mierzinski,  S.     Waterproofing  of  Fabrics.     Trans,  by  A.  Morris 

and  H.  Robson 8vo,     *2  50 

Miller,  E.  H.  Quantitative  Analysis  for  Mining  Engineers ..  8vo,  *i  50 
Miller,  G.  A.  Determinants.  (Science  Series  No.  105.).  .  i6mo, 

Milroy,  M.  E.  W.     Home  Lace -making i2mo,     *i  oo 

Minifie,  W.     Mechanical  Drawing 8vo,     *4  oo 

Mitchell,  C.  A.,  and  Prideaux,  R.  M.     Fibres  Used  in  Textile  and 

Allied  Industries 8vo, 

Modern  Meteorology ' i2mo,       i  50 

Monckton,  C.  C.  F.     Radiotelegraphy.     (Westminster  Series.) 

8vo,     *2  oo 
Monteverde,  R.  D.     Vest  Pocket  Glossary  of  English-Spanish, 

Spanish-English  Technical  Terms 641110,  leather,     *i  oo 

Moore,  E.  C.  S.     New  Tables  for  the  Complete  Solution  of 

Ganguillet  and  Kutter's  Formula 8vo,     *5  oo 

Moreing,  C.  A.,  and  Neal,  T.     New  General  and  Mining  Tele- 
graph Code 8vo,     *s  oo 

Morgan,  A.  P.     Wireless  Telegraph  Construction  for  Amateurs. 

i2mo,     *i  50 


24     D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Moses,  A.  J.  The  Characters  of  Crystals 8vo,  *2  oo 

Moses,  A.  J.,  and  Parsons,  C.  I.  Elements  of  Mineralogy .  .  8vo,  *2  50 
Moss,  S.  A.  Elements  of  Gas  Engine  Design.  (Science 

Series.) i6mo,  o  50 

The  Lay-out  of  Corliss  Valve  Gears.  (Science  Series) .  i6mo,  o  50 

Mullin,  J.  P.  Modern  Moulding  and  Pattern-making.  .  .  .  i2mo,  2  50 
Munby,  A.  E.  Chemistry  and  Physics  of  Building  Materials. 

(Westminster  Series.) 8vo,  *2  oo 

Murphy,  J.  G.  Practical  Mining i6mo,  i  oo 

Murray,.  J.  A.  Soils  and  Manures.  (Westminster  Series.)  .8vo,  *2  oo 

Naquet,  A.     Legal  Chemistry i2mo,  2  oo 

Nasmith,  J.     The  Student's  Cotton  Spinning Svo,  3  oo 

Neilson,  R.  M.     Aeroplane  Patents Svo,  *2  oo 

Nerz,  F.     Searchlights.     Trans,  by  C.  Rodgers Svo,  *3  oo 

Neuberger,   H.,  and  Noalhat,  H.     Technology   of  Petroleum. 

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Newall,   J.  W.      Drawing,  Sizing  and  Cutting  Bevel-gears. 

^                                                                                                     Svo,  i  50 

Newlands,  J.     Carpenters  and  Joiners'  Assistant 

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Nipher,  F.  E.     Theory  of  Magnetic  Measurements i2mo,  i  oo 

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Nolan,  H.     The  Telescope.     (Science  Series  No.  51.) i6mo,  o  50 

Noll,  A.     How  to  Wire  Buildings i2mo,  i  50 

Nugent,  E.     Treatise  on  Optics i2mo,  i  50 

O'Connor,  H.  The  Gas  Engineer's  Pocketbook. .  .  i2mo,  leather,  3  50 

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Ohm,  G.  S.,  and  Lockwood,  T.  D.  Galvanic  Circuit.  Trans,  by 

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Olsson,  A.  Motor  Control,  in  Turret  Turning  and  Gun  Elevating. 

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Parry,  E.  J.     Chemistry  of  Essential  Oils  and  Artificial  Per- 
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Parshall,  H.  F.,  and  Parry,  E.     Electrical  Equipment  of  Tram- 
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Parsons,  S.  J.     Malleable  Cast  Iron 8vo,  *2  50 

Passmore,  A.  C.     Technical  Terms  Used  in  Architecture  . . .  8vo,  *3  50 

Patterson,  D.     The  Color  Printing  of  Carpet  Yarns 8vo,  *3  50 

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Pendred,  V.     The  Railway  Locomotive.     (Westminster  Series.) 

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Perrine,  F.  A.  C.     Conductors  for  Electrical  Distribution  .  .  .  8vo,  *3  50 

Perry,  J.     Applied  Mechanics 8vo,  *2  50 

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Plympton,  G.  W.     The  Aneroid  Barometer.     (Science  Series.) 

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Pullen,  W.  W.  F.  Application  of  Graphic  Methods  to  the  Design 

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Purchase,  W.  R.     Masonry i2mo,  *3  oo 

Putsch,  A.     Gas  and  Coal-dust  Firing 8vo,  *3  oo 

Pynchon,  T.  R.     Introduction  to  Chemical  Physics 8vo,  3  oo 

Rafter,  G.  W.     Mechanics  of  Ventilation.     (Science  Series  No. 

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Raikes,  H.  P.     Sewage  Disposal  Works 8vo,  *4  oo 

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Randall,  P.  M.     Quartz  Operator's  Handbook i2mo,  2  oo 

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Raphael,  F.  C.     Localization  of    Faults  in  Electric  Light  and 

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Rateau,   A.     Flow  of  Steam  through  Nozzles    and    Orifices. 

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Students oblong  4to,  boards,  i  oo 

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Morris  and  H.  Robson i2mo,  *2  50 

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Reynolds,   0.,   and  Idell,   F.   E.     Triple   Expansion   Engines. 

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Rice,  J.  M.,  and  Johnson,  W.  W.     A  New  Method  of  Obtaining 

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Richardson,  J.     The  Modern  Steam  Engine 8vo,  *3  50 

Richardson,  S.  S.     Magnetism  and  Electricity i2mo,  *2  oo 

Rideal,  S.     Glue  and  Glue  Testing 8vo,  *4  oo 

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Robinson,  J.  B.     Architectural  Composition.. ... 8vo,  *2  50 

Robinson,  S.  W.     Practical  Treatise  on  the  Teeth  of  Wheels. 

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Rogers,  A.     A  Laboratory  Guide  of  Industrial  Chemistry.  .  i2mo,  *i  50 

Rogers,  A.,  and  Aubert,  A.  B.     Industrial  Chemistry .  (In  Press.) 

Rogers,  F.     Magnetism  of  Iron  Vessels.     (Science  Series  No.  30.) 

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Rollins,  W.     Notes  on  X-Light 8vo,  5  oo 

Rose,  J.     The  Pattern-makers'  Assistant 8vo,  2  50 

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Ross,  W.  A.     Blowpipe  in  Chemistry  and  Metallurgy.  .  .i2mo,  "*2  oo 
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Rowan,  F.  J.     Practical  Physics  of  the  Modern  Steam-boiler 

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Rowan,  F.  J.,  and  Idell,  F.  E.     Boiler  Incrustation  and  Corro- 
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Roxburgh,  W.     General  Foundry  Practice 8vor  *3  50 

Ruhmer,    E.     Wireless    Telephony.     Trans,    by    J.    Erskine*- 

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Sabine,  R.  History  and  Progress  of  the  Electric  Telegraph.  i2mo,  i  25 

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Sanford,  P.  G.     Nitro-explosives 8vo,  *4  oo 

Saunders,  C.  H.     Handbook  of  Practical  Mechanics i6mo,  i  oo 

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Saunnier,  C.     Watchmaker's  Handbook i2mo,  3  oo 

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Scheele,  C.  W.     Chemical  Essays 8vo,  *2  oo 

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Schmall,  C.  N.     First  Course  in  Analytic  Geometry,  Plane  and 

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Schmall,  C.  N.,  and  Schack,  S.  M.     Elements  of  Plane  Geometry 

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Schumann,  F.     A  Manual  of  Heating  and  Ventilation. 

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Schwartz,  E.  H.  L.     Causal  Geology 8vo,  *2  50 

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Seaton,  A.  E.,  and  Rounthwaite,  H.  M.     Pocket-book  of  Marine 

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Sewell,  T.     Elements  of  Electrical  Engineering 8vo,  *3  oo 

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Sexton,  A.  H.     Fuel  and  Refractory  Materials i2mo,  *2  50 

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Shaw,  P.  E.     Course  of  Practical  Magnetism  and  Electricity. 

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Shields,  J.  E.     Notes  on  Engineering  Construction i2mo,  i  50 

Shock,  W.  H.     Steam  Boilers 4to,  half  mor.,  15  oo 

Shreve,  S.  H.     Strength  of  Bridges  and  Roofs 8vo,  3  50 

Shunk,  W.  F.     The  Field  Engineer i2mo,  mor.,  2  50 

Simmonsv  W.  H.,  and  Appleton,  H.  A.     Handbook  of  Soap 

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Simms,  F.  W.     The  Principles  and  Practice  of  Leveling 8vo,  2  50 

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Sloane,  T.  O'C.     Elementary  Electrical  Calculations i2mo,  *2  oo 

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Smith,  F.  E.     Handbook  of  General  Instruction  for  Mechanics. 

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Soddy,  F.     Radioactivity 8vo,  *3  oo 

Solomon,  M.     Electric  Lamps.     (Westminster  Series.) 8vo,  *2  oo 

Sothern,  J.  W.     The  Marine  Steam  Turbine 8vo,  *s  oo 

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Spang,  H.  W.     A  Practical  Treatise  on  Lightning  Protection 

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Spangenburg,    L.     Fatigue    of   Metals.     Translated   by  S.   H. 

Shreve.     (Science  Series  No.  23.) i6mo,  o  50 

Specht,  G.  J.,  Hardy,  A.  S.,  McMaster,  J.  B.,  and  Walling.     Topo- 
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Speyers,  C.  L.     Text-book  of  Physical  Chemistry 8vo,  *2  25 

Stahl,  A.  W.    Transmission  of  Power.     (Science  Series  No.  28.) 

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Stevenson,  J.  L.     Blast-Furnace  Calculations i2mo,  leather,  *2  oo 

Stewart,  A.     Modern  Polyphase  Machinery i2mo,  *2  oo 

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D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     33 

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Stone,  H.     The  Timbers  of  Commerce 8vo,  3  50 

Stone,  Gen.  R.     New  Roads  and  Road  Laws i2mo,  i  oo 

Stopes,  M.     Ancient  Plants 8vo,  *2  oo 

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Sudborough,  J.  J.,  and  James,  T.  C.     Practical  Organic  Chem- 
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Suffling,  E.  R.  Treatise  on  the  Art  of  Glass  Painting 8vo,  *3  50 

Swan,  K.  Patents,  Designs  and  Trade  Marks.  (Westminster 

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Sweet,  S.  H.  Special  Report  on  Coal 8vo,  3  oo 

Swinburne,  J.,  Wordingham,  C.  H.,  and  Martin,  T.  C.  Electric 

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Swoope,  C.  W.  Practical  Lessons  in  Electricity i2mo,  *2  oo 

Tailfer,  L.     Bleaching  Linen  and  Cotton  Yarn  and  Fabrics .  .  8vo,  *5  oo 
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Thompson,  E.  P.     How  to  Make  Inventions 8vo,  o  50 

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Thompson,  W.  P.     Handbook  of  Patent  Law  of  All  Countries 

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Tinney,  W.  H.     Gold-mining  Machinery 8vo,  *3  oo 

Titherley,  A.  W.     Laboratory  Course   of    Organic  Chemistry 

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Tonge,  J.     Coal.     (Westminster  Series.) 8vo,  *2  oo 

Townsend,  F.     Alternating  Current  Engineering —  8vo,  boards,  *o  75 

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Trowbridge,  W.  P.     Turbine  Wheels.     (Science  Series  No.  44.) 

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Tucker,  J.  H.     A  Manual  of  Sugar  Analysis 8vo,  3  50 

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Tunner,   P.    A.     Treatise    on   Roll-turning.     Trans,    by   J.    B. 

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Turbayne,  A*.'  A.     Alphabets  and  Numerals 4to,  2  oo 

Turnbull,  Jr.,  J.,  and  Robinson,  S.  W.     A  Treatise  on  the  Com- 
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Turrill,  S.  M.     Elementary  Course  in  Perspective i2ino,  *i  25 

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Vega,  Baron,  Von  Logarithmic  Tables 8vo,  half  mor.,  2  50 

Villon,  A.  M.  Practical  Treatise  on  the  Leather  Industry. 

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Volk,  C.  Haulage  and  Winding  Appliances 8vo,  *4  oo 

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Wabner,  R.     Ventilation  in  Mines.     Trans,  by  C.  Salter.  .  .8vo,  *4  50 

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Walker,  S.  F.     Steam  Boilers,  Engines  and  Turbines 8vo,  3  oo 

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Wood,  De  V.     Luminiferous  Aether.     (Science  Series  No.  85.) 

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